WBBSE Solutions For Class 9 Physical Science Chapter 2 Force And Motion Topic A Rest Motion And Equation Of Motion

Chapter 2 Topic A Rest Motion And Equation Of Motion Synopsis

The state of rest or motion of a body is always observed with respect to some other bodies in the surrounding. This some other bodies provide the frame of reference or reference frame.

There are two types of reference frame:

1. Inertial frame and
2. Non-inertial frame.

If a body does not change its position with respect to a neighboring object with the change of time, then the body is said to be a static body and its state is known as state of rest.

If a body changes its position with respect to a neighbouring object with change of time, then the body is said to be a dynamic body and its state is known as state of motion.

Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment

The motion of a particle is generally of two types:

1. Translational or linear motion and
2. Rotating motion.

Translational motion is the motion of a body along a straight line.

If a body rotates around a fixed axis or a fixed point, its motion is called rotational motion. If a body undergoing periodic motion moves repeatedly along the same straight line at equal intervals of time, then its motion is called rectilinear oscillatory motion.

If a particle rotates around an axis or a point in a circular path, its motion is said to be a circular motion.

Speed of a moving particle is the distance traversed by it in unit time. Speed is a scalar quantity. Unit of speed in SI is m/s. Dimensional formula of speed is LT^-1.

If a moving particle traverses equal distances in equal time intervals, its corresponding speed is said to be uniform speed.

If a moving particle traverses different distances in equal time intervals, its corresponding speed is said to be non-uniform.

For a particle moving with non-uniform speed, its average speed is obtained by dividing the total distance traversed by the total time required to traverse that distance.

If a moving particle traverses s₁ distance in the first t₁ time, s₂ distance in the next t₁ time, and s₃ distance in the last t₃ time, then. average speed,

\(v_a=\frac{s_1+s_2+s_3}{t_1+t_2+t_3}\)

If the magnitude and direction of the velocity of a particle always remain unchanged with time, then the velocity of that particle is called uniform velocity.

If the magnitude and direction or both of a particle change with respect to time, then the velocity of that particle is called non-uniform velocity.

class 9 chapter 2 physics

If a particle moves along a circular path with uniform speed, then the motion of the particle is called uniform circular motion.

It is the motion with uniform speed. In this case, acceleration of the particle (centripetal acceleration) acts towards the centre along the radius of the circular path.

Rate of change of velocity of a particle with respect to time is called its acceleration. Since acceleration has both magnitude and direction, hence it is a vector quantity. Unit of acceleration in SI is m/s² and the dimensional formula is LT^-2.

The initial velocity of a particle moving in a straight line is u. If its velocity is v after time t, then its acceleration, a = \(\frac{v-u}{t}\).

Deceleration is negative acceleration.

The directions of velocity and acceleration are the same if the velocity of a particle moving in a straight line increases but if the velocity of the particle decreases, velocity and acceleration are in the opposite directions.

Equations Of Dynamics: If initial velocity of a particle moving in a straight line = u, uniform acceleration = a, velocity after time t = v, and distance traversed in time t = s, then,

1. \(v=u-a t\)
2. \(s=u t-\frac{1}{2} a t^2\)
3. \(v^2=u^2-2 a s\)

Case 1: If the initial velocity of the particle, u = 0, that is, the particle started its journey from rest,

1. v = at
2. s = 1/2 at²
3. v² = 2as

Case 2: If the particle is moving with uniform deceleration of a, starting with initial velocity of u,

1. v = u – at
2. s = ut- \(/frac{1}{2}\)at²
3. v² = u² – 2as

Case 3: If the particle falls under gravity with an initial velocity of u, it comes down with an uniform acceleration of a = g. Here, if s = h,

1. \(v=u+g t\)
2. \(h=u t+\frac{1}{2} g t^2\)
3. \(v^2=u^2+2 g h\)

Case 4: If the particle falls from rest under gravity,

1. \(v=g t\)
2. \(h=\frac{1}{2} g t^2\)
3. \(v^2=2 g h\)

Case 5: If the particle is thrown upwards in a perpendicular direction with u velocity,

1. \(v=u-g t\)
2. \(h=u t-\frac{1}{2} g t^2\)
3. \(v^2=u^2-2 g h\)

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Chapter 2 Topic A Rest Motion And Equation Of Motion Short And Long Answer Type Questions

Question 1. What do you understand by rest and motion?

Answer:

Rest: If a body does not change its position with change of time and with respect to a object, then that body is said to be a stationary body. In this condition, it is said to be at rest.

Motion: If a body changes its position with change of time and with respect to a neighbouring object, then that body is said to be a moving body. In this condition, it is said to be in motion.

Question 2. All rest and motion are relative- explain.

Answer:

All objects like houses, trees, etc. surrounding us do not change their positions with respect to the earth’s surface and also with change of time. So these are called stationary objects with respect to the surface of the earth.

But the earth itself is moving around the sun. So, houses, trees, etc. which are stationary with respect to the earth are not at absolute rest. In reality, every object of this universe is in motion with respect to every other object.

So, there is no state of absolute rest in nature. Since there is no existence of absolutely stationary bodies, so there is also no absolutely moving body. Thus, all rest and motion are relative.

Key Questions on Rest and Motion for Class 9

Question 3. How many types of motion a body may possess? What are those types?

Answer:

A body may possess two types of motion.

These Two Types Of Motion Are

1. Rotational motion and
2. Translational motion or linear motion.

Question 4. What do you understand by rotational motion? Explain with examples.

Answer:

If a body rotates with respect to one of its axes, then the motion of the body is called rotational motion.

Example: The motion of blades of a running electric fan is rotational motion. Again, the diurnal motion of the earth around its axis throught day and night like a top is also a rotational motion.

“numericals of physics class 9 chapter 2 “

Question 5. What is translational motion? Explain with examples.

Answer:

If a body moves in a straight line, its motion is called translational motion.

Example: If a stone is dropped from a roof top, it falls straight downward. In this case, motion of the stone is a translational motion.

Practice Questions for Chapter 2 Rest and Motion

Question 6. What is compound motion? Give example.

Answer:

If a body is moving in such a way that its motion is neither a pure translational motion nor a rotational motion but a mixture of both, then this motion is called a compound motion.

Example: Let us consider the motion of wheel of a running car. Except the particle at the centre point of the wheel, all other particles are undergoing compound motion.

Question 7. Write down the differences between translational motion and rotational motion.

Answer:

Differences between translational motion and rotational motion:

Question 8. What do you mean by rotational plane and axis of rotation?

Answer:

Rotational Plane: When a particle rotates around a fixed axis or point, then during the state of rotation the particle makes a circle. The horizontal plane on which the circle lies is called rotational plane.

Axis Of Rotation: When a particle rotates in a circular path, the straight line perpendicular to the rotational plane and passing through the centre of the circle is called the axis of rotation.

Question 9. What is the difference between a rotational motion and a circular motion? Explain with examples.

Answer:

A rigid body consists of more than one particle. When such a rigid body rotates around a fixed axis of its own, then this motion is called rotational motion.

In this case, the body does not change its place with respect to its axis but simply changes its orientation. The axis about which the body
rotates is called the rotational axis.

The motion of the blades of a fan is rotational motion. In this case, the blades rotate with respect to the axis situated at the centre of the fan.

In the same way, the motion of a top remaining stationary at one place on the floor, motion of a disc on CD drive, dlurnal motion of the earth (due to which the phenomenon of day and night occurs) etc. are examples of rotational motion.

“force and law of motion class 9 question answer ”

On the other hand, the motion of a particle along a circular path is the circular motion of the particle. Since the particle has the shape of a point, hence it is not possible for the particle to have an axis. So the particle has no rotational motion.

The revolving motion of an electron around the nucleus is circular motion. Again, if the earth is considered to be a point in comparison to the sun, its motion around the sun (due to which seasonal changes occur) is the earth’s circular motion.

Question 10. Diurnal and annual motion of the earth fall under which category?

Answer:

If we assume that the earth rotates in a circular path around the sun (actually it is an elliptical path), annual motion of the earth is a circular motion.

While moving around the sun, the earth rotates with respect to its own axis. This is called diurnal motion which falls under rotational motion.

Question 11. What do you understand by periodic motion? Give examples.

Answer:

If a body traverses the same path repeatedly after equal intervals of time, then the motion of the body is known as periodic motion.

Example: Diurnal motion of the earth around the sun, motion of the hand of a clock, motion of a simple pendulum etc. are examples of periodic motion.

Question 12. What is displacement? What type of quantity is it?

Answer: The change of position of a moving body in a particular direction is the displacement of the body. Displacement is a vector quantity because it has both magnitude and direction.

Question 13. Can the displacement of a particle be greater than the traversed distance?

Answer:

The displacement of a moving particle can never be greater than the traversed distance. If the particle travels from one point to another in a straight line path, the value of displacement becomes equal to that of the traversed distance but if the particle moves in a curved path, value of displacement is less than that by distance traversed.

Question 14. What are the differences between displacement and distance traversed?

Answer:

Differences between displacement and distance traversed:

Question 15. What is speed? What type of quantity is speed? What are the units of speed in CGS system and SI?

Answer:

1. The amount of distance a moving particle traverses in unit time is called its speed.
2. Speed has magnitude but no direction, hence it is a scalar quantity.
3. Units of speed in CGS system and SI are cm/s and m/s respectively.

Question 16. Define uniform speed and non-uniform speed.

Answer:

Uniform Speed: A moving particle has a uniform speed if it travels equal distances in equal intervals of time.

Non-Uniform Speed: A moving particle has a non-uniform speed if it travels unequal distances in equal intervals of time.

Question 17. What do you mean by average speed?

Answer:

Average speed is the ratio of the distance traversed and the time taken to cover that distance by a moving particle traveling with non-uniform speed.

Suppose, while going to your friend’s house, you cover the first s₁ distance by rickshaw in time t₁, the next s₂ distance by auto-rickshaw in time t_2, and the last s₃ distance by walking in time t₃. In this case, you cover the whole journey with non-uniform speed.

Here, the average speed is defined as

\(v_a=\frac{s_1+s_2+s_3}{t_1+t_2+t_3}\)

Question 18. How do you define velocity? What type of quantity is velocity?

Answer:

1. Rate of change of displacement of a body with respect to time in a particular direction is called the velocity of a body.
2. Velocity is a vector quantity because it has both moving in a circular path, the velocity of the magnitude and direction.

Important Concepts in Equations of Motion for Class 9

Question 19. Define uniform velocity and non-uniform velocity.

Answer:

Uniform Velocity: A particle is said to have uniform velocity if it covers equal distance in equal intervals of time in a given direction.

Non-Uniform Velocity: A particle is said to have non-uniform velocity if the magnitude or direction or both of the particle change with respect to time.

“force and laws of motion exercise ”

Question 20. Write down the differences between speed and velocity.

Answer:

Differences between speed and velocity:

Question 21. Explain how a body with uniform speed may not have uniform velocity.

Answer:

A body is said to have uniform speed if it traverses equal distances in equal intervals of time in a curved path. As the body is moving in a curved path, the direction of its velocity changes every moment so it is not moving with uniform velocity.

Hence, a body moving with uniform speed need not have uniform velocity, if it traverses equal distances in equal intervals of time by changing its direction.

Question 22. What do you mean by uniform circular motion?

Answer:

Suppose, a particle is moving in a circular path with uniform speed as shown in the figure. While particle at any moment at a point on the circumference always acts in a tangent to that point.

As a result, the velocity of the particle changes every moment. This type of motion of a particle is called uniform circular motion.

Question 23. What do you mean by acceleration?

Answer:

The rate of change of velocity of a moving particle with respect to time is called its acceleration.

Suppose the initial velocity of a particle moving in a straight line = u and after time t, its velocity becomes v.

∴ according to the definition of acceleration of the particle, \(a=\frac{v-u}{t}\)

Question 24. Determine the dimensional formula of acceleration. What is the dimension of acceleration? What are the units of acceleration in CGS system and SI?

Answer:

Dimensional formula of acceleration

= \(\frac{\text { dimensional formula of velocity }}{\text { dimensional formula of time }}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^{-1}}=\mathrm{LT}^{-2}\)

Dimension of acceleration is 1 in length and -2 in time.

Units of acceleration in CGS system and SI are cm/s² and m/s², respectively.

Question 25. Displacements of a particle in time t, s = 4t + 6t². If displacement is expressed in unit m and time in unit s, what is the acceleration of the particle?

Answer:

If we compare the two equations, s = 4t + 6t² and s = ut + \(\frac{1}{2}\)at²,

we find, \(\frac{a}{2}\) = 6 or, a = 12 m/s²

∴ acceleration of the particle = 12 m/s².

Question 26. What are uniform acceleration and non-uniform acceleration?

Answer:

Uniform Acceleration: A moving particle is said to have uniform acceleration if it travels in a straight line and its velocity increases by equal amount in equal intervals of time.

Non-Uniform Acceleration: A moving particle is said to have non-uniform acceleration if its velocity increases by unequal amounts in equal intervals of time.

Question 27. A moving particle starting from rest traverses a distance x in 1 s with uniform acceleration. How much distance does it cover in 2 s time?

Answer:

Distance traversed by a particle moving with uniform acceleration starting from rest in time t, s ∝ t².

∴ The particle covers a distance of 4x in 2 s time.

Question 28. If the initial velocity of a moving particle with uniform acceleration is zero, what is the nature of its velocity-time curve?

Answer:

The graph of a moving particle with uniform acceleration is always a straight line. Again, since the initial velocity of the particle is zero, hence the graph will always pass through the origin.

Question 29. What is instantaneous acceleration?

Answer: Instantaneous acceleration is the limiting value of the ration of the change in velocity of a moving particle during a very small change of time at any particular instant.

Understanding Types of Motion for Solutions

Question 30. Is it possible for a moving particle to have acceleration but constant magnitude of its velocity?

Answer:

If a particle moves in such a way that its acceleration always acts perpendicularly to the direction of its motion, the magnitude of velocity remains unchanged even though the moving particle has acceleration.

For example, the acceleration of a particle moving in a circle with uniform speed is acting towards the centre along the radius of the circle and at any moment, velo- city is acting at that point along the tangent to the circle.

In this case, angle between acceleration and velocity is always 90°. So for this type of motion, magnitude of velocity always remains unchanged even though acceleration is present.

Question 31. Deceleration is negative acceleration- explain.

Answer:

If the velocity of a moving particle reduces with the change of time, then the rate of change of velocity or acceleration is negative. This negative acceleration is called deceleration or retardation.

In this case, deceleration and velocity act in opposite directions.

Question 32. Why is the phrase ‘per second’ used twice in the unit of acceleration?

Answer:

The word second is used twice in the unit of acceleration because it has two significances. It signifies change of velocity as well as rate of change of velocity.

Question 33. Is it possible for a moving particle to have velocity but no acceleration?

Answer:

It is possible for a particle moving with uniform velocity to have velocity but no acceleration. If a particle remains stationary or moves with uniform velocity, it does not have any acceleration.

“laws of motion class 9th exercise “

Question 34. Is it possible for a body to have acceleration but no velocity?

Answer: It is possible that a particle has acceleration but no velocity. A body is thrown upwards in a perpendicular direction. It moves up and at the highest point, it comes to rest for a moment.

At that time, velocity is zero but acceleration due to gravity still works on the body.

Question 35. Establish the equation s = ut+1/2at² by algebraic method.

Answer: Let the initial velocity of a particle moving in a straight line with uniform acceleration a and its velocity after time t be u and v, respectively.

∴ v = u + at

Since the particle is moving with uniform 0.38 Show that the average velocity of a acceleration, rate of increase of velocity is the same.

∴ Average velocity of the particle at time t,

\(v_a=\frac{u+v}{2}\)

As the particle is moving with uniform acceleration, distance traversed by the particle at time t is equal to the distance traversed in time t with average velocity V_a.

If s is the distance traversed by the particle in time t, then

s = \(v_a \cdot t=\frac{u+v}{2} \cdot t=\frac{u+u+a t}{2} \cdot t\)

= \(\frac{(2 u+a t) t}{2}=\frac{2 u t+a t^2}{2}\)

∴ \(s=u t+\frac{1}{2} a t^2\)

Question 36. Establish the equation v² = u² + 2as by algebraic method.

Answer:

Let the initial velocity of a particle moving in a starigh line with uniform acceleration a and velocity after time t be u and v respectively.

∴ v = u + at ……(1)

If s is the distance traversed by the particle in time t, then

s = ut + \(\frac{1}{2}\)at² …..(2)

Squaring both sides of euqation (1), we get

\(v^2=(u+a t)^2=u^2+2 u a t+a^2 t^2\)

= \(u^2+2 a\left(u t+\frac{1}{2} a t^2\right)\)

Now from equation (2), we get

v² = u² + 2as

Question 37. The equation \(s=u t+\frac{1}{2} a t^2+s_0\) signifies what type of motion?

Answer:

If we put t=0 in the equation \(s=u t+\frac{1}{2} a t^2+s_0\), we get s = S₀

∴ The particle starts its movement from position s = s₀ with initial velocity u and uniform acceleration a.

Question 38. Show that the average velocity of a particle in case of motion with uniform acceleration is half of the sum of initial and final velocities.

Answer: Let the initial velocity of a particle moving with uniform acceleration be u and its velocity after time t be v.

Distance traversed by the particle in time t,

\(s=u t+\frac{1}{2} a t^2\)

∴ Average Velocity,

\(v_a =\frac{s}{t}=\frac{u t+\frac{1}{2} a t^2}{t}=u+\frac{1}{2} a t\)

= \(\frac{2 u+a t}{2}=\frac{u+u+a t}{2}=\frac{u+v}{2}\)

Question 39. The initial velocity of a particle moving with uniform acceleration is u and after travelling a distance s, its velocity becomes v. What is its velocity after it covers half of the total path?

Answer:

If the acceleration of the particle is a, then v² = u² + 2as.

Let the velocity of the particle is v₁ after it traverses half of the total distance or \(\frac{s}{2}\).

∴ \(v_1^2=u^2+2 a \cdot \frac{s}{2}=\frac{2 u^2+2 a s}{2}\)

= \(\frac{u^2+u^2+2 a s}{2}=\frac{u^2+v^2}{2}\) \([because v^2=u^2+2 a s]\)

∴ velocity of the particle after it traverses distance \(\frac{s}{2}\),

\(v_1=\sqrt{\frac{u^2+v^2}{2}}\)

Question 40. Initial velocity of a moving particle with uniform acceleration is u, after time tits velocity is v. What is the velocity of the particle after half the time \(\frac{t}{2}\)?

Answer:

If the velocity of the particle is a, then v = u + at.

Velocity after time \(\frac{t}{2}\),

\(v_1=u+\frac{a t}{2}=\frac{2 u+a t}{2}=\frac{u+u+a t}{2}=\frac{u+v}{2}\) \([because v=u+a t]\)

Sample Solutions from WBBSE Class 9 Physical Science Chapter 2

Question 41. 41 Distances traversed by a particle moving with uniform acceleration and starting from rest in the first, second, and third seconds are s₁, s₂, and s₃, respectively. Determine s₁ : s₂ : s₃.

Answer: 

If the acceleration of the particle is a, then \(s_1=\frac{1}{2} a \times 1^2=\frac{a}{2}\)

If the particle traverses a distance s’₂ in 2s, then \(s_2^{\prime}=\frac{1}{2} a \times 2^2=2 a\)

∴ \(s_2=s_2^{\prime}-s_1=2 a-\frac{a}{2}=\frac{3}{2} a\)

Again, if the particle traverses a distance s’₃ in 3s, then

\(s_3^{\prime}=\frac{1}{2} \times a \times 3^2=\frac{9}{2} a\)

∴ \(s_3=s_3^{\prime}-s_2^{\prime}=\frac{9}{2} a-2 a=\frac{5}{2} a\)

So, \(s_1: s_2: s_3=\frac{a}{2}: \frac{3}{2} a: \frac{5}{2} a=1: 3: 5\)

Question 42. Establish the equation v=u+at with the help of a velocity-time graph.

Answer:

Let the initial velocity of a particle moving with uniform acceleration along a straight line be u and after time t, let its velocity be v.

OX and OY are two mutually perpendicular axes.

A velocity-time line graph of the particle is drawn by taking time (t) along x-axis and velocity (v) along y-axis. Straight line AB represents the velocity-time line graph of the particle.

Here, initial velocity, OA = u and at any definite time, OC=t, velocity of the particle, BC = v. A straight line AD is drawn parallel to OC.

From the graph, we may write BC = BD + DC = BD + OA [because OA=DC]

∴ v = BD + u or, BD = v – u …(1)

As per the graph, acceleration of the particle,

\(a=\frac{\text { change of velocity }}{\text { time }}=\frac{B C-A C}{O C}\)

= \(\frac{B C-D C}{O C}=\frac{B D}{O C}=\frac{B D}{t}\) [because OC = t]

∴ BD = at…(2)

By comparing equations (1) and (2), we get at=v-u or, v=u+at

Question 43. Establish the equation s = ut + \(\frac{1}{2}\) at² with the help of a velocity-time graph.

Answer:

Let a particle with initial velocity u is moving along a straight line with uniform acceleration a. Now suppose v is the velocity of the particle after time t and distance traversed by the particle during that time is s.

OX and OY are two mutually perpendicular axes. If time is expressed along x- axis and velocity is expressed along y-axis, straight line AE expresses velocity-time graph of the particle.

Here, OA = u is the initial velocity of the particle. Straight line AF parallel to the time axis and passing through the point A expresses velocity-time graph of the particle in a state without acceleration.

At any specific time, OB = t, velocity of the particle, BC = v = u + at.

Distance traversed by the particle in time t,

s = area of trapezium OACB

= \(\frac{1}{2}\)(OA + BC) X OB

∴ s = \(\frac{1}{2}\)(u+u+at) xt

= \(\frac{1}{2}\)(2ut + at²)

= \(\frac{1}{2}\)ut + at²

Question 44. Establish the relationship s=vt by using velocity-time graph.

Answer: Suppose, a particle is moving with uniform velocity v.

OX and OY are two mutually perpendicular axes. If time is represented by x- axis and velocity by y-axis, then length OA represents uniform velocity v of the particle.

As the particle is moving with uniform velocity, there would not be any change in its velocity with change of time.

So, velocity-time graph is a Istraight line AB parallel to the time axis. Suppose, length OC represents a definite time t. A perpendicular CD is drawn from C to AB.

Distance traversed by the particle in time t, s = vt = OA x OC = area of the rectangle OADC.

Hence, the area formed by the velocity-time graph with the time axis in a definite time gives the distance traversed by the particle.

Question 45. Establish the equation s = \(\frac{1}{2}\) at² with the help of a velocity-time graph.

Answer:

Let a particle starting from rest, travels along a straight line with uniform acceleration and after time t, its velocity is v.

OX and OY are two mutually perpendicular axes. If time is expressed along x-axis and velocity is expressed along y-axis, straight line OA which passes through the point of origin will express velocity- time graph of the particle.

According to the line graph, velocity of the particle, BC = v at any definite time OB = t.

Distance traversed by the particle at time t,

s = area of ΔOBC

= \(\frac{1}{2}\) x OB x BC = \(\frac{1}{2}\) x t x v = \(\frac{1}{2}\) at²

Question 46. Establish the equation v²=u²+2as with the help of a velocity-time graph.

Answer:

Let a particle with initial velocity u is moving along a straight line with uniform acceleration a.

Now suppose v is the velocity of the particle after time t and distance traversed by the particle during that time is s.

OX and OY are two mutually perpendicular axes. If time is expressed along x-axis and velocity is exposed along y-axis, straight line AE expresses velocity-time graph of the particle.

Here, OA = u is the initial velocity of the particle. Straight line AF parallel to the time axis and passing through the point A expresses velocity- time graph of the particle in a state without acceleration.

At any specific time, OB =t, velocity of the particle, BC = v = u + at.

Distance traversed by the particle in time t,

s = area of trapezium OACB

= \(\frac{1}{2}\)(OA + BC) X OB

= \(\frac{1}{2}\)(OA + BC) X OB X \(\frac{CD}{CD}\)

= \(\frac{1}{2}\)(OA + BC) x OB x \(\frac{BC-BD}{CD}\)

= \(\frac{1}{2}\)(OA + BC) x (BC -OA) x \(\frac{OB}{CD}\)

= \(\frac{1}{2}\)(u + v) x (v – u) x \(\frac{t}{at}\)

[because CD = BC = v – u = at]

or, \(s=\frac{v^2-u^2}{2 a}\)

or, \(v^2-u^2=2 a s\)

∴ \(v^2=u^2+2 a s\)

Question 47. Draw the displacement-time graph of a particle moving with uniform velocity. What is the nature of this graph?

Answer:

Suppose, a particle is moving with uniform velocity v.

∴ Distance traversed by the particle in time t, s = vt

If time (t) is expressed along x-axis and displacement is expressed along y-axis in the line graph, s-t graph (OA) of the particle is a straight line passing through the origin.

Question 48. Draw the displacement-time graph of a particle moving with uniform acceleration. What is the nature of this line graph?

Answer:

Suppose, a particle is moving with uniform acceleration a. If the initial velocity of the particle is u, distance traversed in time t,

s = ut + \(\frac{1}{2}\) at²……(1)

In the line graph, if time (t) is expressed along x-axis and displacement (s) is expressed along y-axis, s-t graph (OA) of the particle will be a parabola passing through the origin.

Question 49. Velocity-time graphs of four different particles are shown in the given figure. Out of these, no. 1 and no. 2 curves are parallel to each other. Answer the following questions according to the image:

1. Which particles have zero initial velocity?
2. Which particle is moving with uniform velocity?
3. Which particles have same acceleration?
4. Among the particles moving with uniform acceleration, which particle or particles have lowest acceleration?

Answer:

1. Curves no. 2 and no. 3 are passing through the origin. So the initial velocity of both the particles is zero.
2. Curve no. 4 curve is parallel to the time axis. So the corresponding particle is moving with uniform velocity.
3. Curves no. 1 and no. 2 are parallel to each other. Hence the corresponding particles have the same acceleration.
4. Out of the curves numbered 1, 2, 3, and 4, the curve no. 3 is inclined to the x-axis with lowest angle. Hence, the acceleration of the particle corresponds to curve no. 3 is the lowest.

Question 50. Velocity-time graphs of two bikes in a bike competition are shown in the image. At the end of this competition of two hours duration:

 

1. Which bike continues to have greater velocity?
2. Which bike traverses greater distance?
3. Which bike has the greater acceleration?

Answer:

It is seen that after two hours, both the graphs meet at the same point. That is, the velocities of both particles are the same and that is 80 km/h.

According to the graphs, the curve corresponding to the first bike is line OAB and for the second bike, it is line OB.
In this case, area of quadrilateral OABC > area of triangle OBC.

In the velocity-time graph, the area contained within the curve and the time axis at a definite time indicates the distance traversed by the concerned particle. Hence, the first bike covers a greater distance.

The first bike traveled the first one hour with uniform acceleration and the second bike was in motion with uniform acceleration throughout the journey. Since the curve OA is inclined to the x-axis with a greater angle than the curve OB, the acceleration of the first car is greater.

Chapter 2 Topic A Rest Motion And Equation Of Motion Multiple Choice Questions

Question 1. A ball is thrown. What type of motion will the ball follow if rotation of the ball along its own axis is ignored?

1. Compound motion
2. Circular motion
3. Linear motion
4. Oscillatory motion in a straight line

Answer: 1. Compound motion

Question 2. A particle traverses a semi-circular path of 1m radius in 1s time. What is the average velocity of the particle?

1. 3.14 m/s
2. 2 m/s
3. 1 m/s
4. Zero

Answer: 2. 2 m/s

Question 3. Starting from your house, you walk 8 km in 2 hours and come back to the house. What will be your displacement?

1. 4 km
2. 2 km
3. 8 km
4. Zero

Answer: 4. Zero

Question 4. n number of bullets are pumped out per second from a machine gun. If mass of each bullet is m kg and its velocity is v m/s, what is the applied force (in N unit) on this machine gun?

1. mnv
2. \(\frac{mn}{v}\)
3. mn
4. \(\frac{mv}{n}\)

Answer: 1. mnv

Question 5. An example of compound motion is

1. A ball rolling on the ground
2. Rotation of a ball at one place
3. Slippery motion of a ball
4. Perpendicular fall of a ball from a height

Answer: 1. A ball rolling on the ground

Question 6. A plane flies 6000 km eastward and then 8000 km northward. Then the plane comes back to its initial position by the shortest route. If the speed of this plane is 200 km/h, what was its average velocity in the total journey path?

1. 0
2. 120 km/h
3. 200 km/h
4. 220 km/h

Answer: 1. 0

Question 7. Which of the following quantity remains unchanged in rotational motion?

1. Velocity
2. Axis
3. Linear momentum
4. None of these

Answer: 2. Axis

Question 8. The motion of a wheel of a running car is

1. Translational motion
2. Rotational motion
3. Compound motion
4. None of the above

Answer: 3. Compound motion

Question 9. Ratio of traversed distance and displacement of a moving body is

1. <1
2. ≤1
3. ≥1
4. = 1

Answer: 3. ≥1

Question 10. A man travels along the circumference of a semi-circular field of radius 14 m and goes to the other side. The magnitude of displacement of the man is

44 m
28 m
88 m
14 m

Answer: 2. 28 m

Question 11. A particle moves from point A along the circumference of a circle of radius 5√2 cm to point B, so that an angle of 60° is formed at the centre. The displacement of the particle is

1. 5√2 cm
2. 5 cm
3. 10√2 cm
4. 10 cm

Answer: 1. 5√2 cm

Question 12. Initial velocity and deceleration of a particle are 20 m/s and 2.5 m/s² respectively. Time taken by the particle to come to rest is

1. 4s
2. 6s
3. 8s
4. 10s

Answer: 3. 8s

Question 13. For uniformly circular motion

1. Velocity of a particle always remains unchanged
2. Speed and acceleration of a particle remain unchanged
3. Speed changes
4. Magnitudes of speed and acceleration remain unchanged

Answer: 4. Magnitudes of speed and acceleration remain unchanged

Question 14. Velocity-time (v-t) graph of a particle. Average velocity of the particle is

1. 4 m/s
2. 5 m/s
3. 6 m/s
4. 7.5 m/s

Answer: 2. 5 m/s

Question 15. Velocity-time (v-t) graph of a particle is shown in the figure. Acceleration of the particle is

1. 0.2 m/s²
2. 0.3 m/s²
3. 0.4 m/s²
4. 0.5 m/s²

Answer: 3. 0.4 m/s²

Question 16. Dimensional formula of acceleration is

1. LT^-2
2. LT^-2
3. LT^-3
4. L^-1T

Answer: 2. LT^-2

Question 17. A particle moving with uniform velocity has an initial velocity u and a final velocity v. The average velocity of the particle is

1. \(\frac{v-u}{2}\)
2. \(\frac{u+v}{2}\)
3. \(\sqrt{\frac{u^2+v^2}{2}}\)
4. \(\sqrt{\frac{v^2-u^2}{2}}\)

Answer: 2. \(\frac{u+v}{2}\)

Question 18. Velocity-time (v-t) graph of a particle is shown in the figure. Ratio of distances traversed by the particle with uniform acceleration and with uniform velocity is

1. 1:2
2. 2:1
3. 2:3
4. 3:5

Answer: 1. 1:2

Question 19. Which of the following is at absolute rest?

1. Earth
2. Sun
3. Moon
4. None of these

Answer: 4. None of these

Question 20. Motion of all moving bodies is

1. Absolute motion
2. Relative motion
3. Translational motion
4. Rotational motion

Answer: 4. Rotational motion

Question 21. Example of translational motion is

1. Rolling motion of a ball on ground
2. Vertical fall of a ball from a height
3. Rotation of a ball at one place
4. Slippery motion of a ball

Answer: 2. Vertical fall of a ball from a height

Question 22. Direction of motion of particles inside a body which moves with translational motion in a straight line

1. Always remains unchanged
2. Always changes
3. May remain unchanged ormay change
4. Changes at first and then remains unchanged

Answer: 1. Always remains unchanged

Question 23. Direction of motion of particles inside a body which moves with oscillatory motion in a straight line

1. Always remains unchanged
2. Changes at an interval of time
3. Always changes
4. None of the above

Answer: 2. Changes at an interval of time

Question 24. Rest and motion of a body is measured in comparison with another body which is called the

1. Reference frame
2. Relative frame
3. Body at rest
4. Reference body

Answer: 4. Reference body

Question 25. Unit of acceleration in CGS system is

1. m.s^-1
2. cm.s^-2
3. m.s^-2
4. cm.s^-1

Answer: 2. cm.s^-2

Question 26. To express speed

1. Only magnitude is required
2. Only direction is required
3. Both magnitude and direction are required
4. Either magnitude or direction is required

Answer: 1. Only magnitude is required

Question 27. Velocity and acceleration of a body moving in a straight line

1. Always act in the same direction
2. Always act in opposite directions
3. Act in opposite directions in some cases
4. Never act in opposite directions

Answer: 1. Always act in the same direction

Question 28. Which of the following is a constant quantity in case of a freely falling body?

1. Displacement
2. Acceleration
3. Velocity
4. Speed

Answer: 2. Acceleration

Question 29. A particle moves half the distance of a semi- circular path with radius r. Displacement of this particle is

1. r
2. 2r
3. r/2
4. 3r

Answer: 2. 2r

Question 30. The initial velocity and acceleration of a particle are u and a, respectively. After time t, its velocity is v. The relationship between u, v, a, and t is given by

1. u = v + at
2. u + v = at
3. v – u = at
4. v = u + at

Answer: 4. v = u + at

Question 31. A boy walks 3 km eastwards and then 4 km northwards. Displacement of the boy is

1. 10 km
2. 5 km
3. 8 km
4. 14 km

Answer: 2. 5 km

Question 32. Dimensional formula of velocity is

1. LT^-1
2. LT^-2
3. LT
4. MLT^-1

Answer: 1. LT^-1

Question 33. A body at rest starts moving with uniform acceleration of 8 cm s^-2 when a force is applied on it. Its final velocity after 12s is

1. 50 cm.s^-1
2. 75 cm-s^-1
3. 80 cm. s^-1
4. 96 cm.s^-1

Answer: 4. 96 cm.s^-1

Question 34. Area between the curve and time axis in a velocity-time graph indicates

1. Displacement of the body
2. Acceleration of the body
3. Change of velocity of the body
4. None of the above

Answer: 1. Displacement of the body

Question 35. A ball is thrown by a man upward in a perpendicular direction with a velocity of 10m s^-1. After sometime, the ball comes back to the man again. Average velocity of the ball is

1. 15 m.s^-1
2. 10 m.s^-1
3. 20 m.s^-1
4. Zero

Answer: 4. Zero

Question 36. Which one of the four graphs shown below represents a body moving with uniform acceleration?

Answer: 2.

Question 37. The velocity-time graph of a particle moving with uniform velocity is

1. A straight line parallel to the time axis
2. A straight line parallel to the velocity axis
3. A straight line passing through the origin
4. A curved line passing through the origin

Answer: 1. A straight line parallel to the time axis

Question 38. A boy is tossing up a coin perpendicularly inside a running compartment of a train with his face towards the engine. If the coin falls behind the boy, then the train

1. Is moving forward with uniform velocity
2. Is moving backward with uniform velocity
3. Is moving forward with deceleration
4. Is moving forward with acceleration

Answer: 4. Is moving forward with acceleration

Question 39. In a rotational motion

1. Velocity of any particle is constant
2. Velocity of a particle further from the axis of rotation is greater
3. Velocity of a particle further from the axis of rotation is lesser
4. Angular velocity and rectilinear velocity are constant

Answer: 2. Velocity of a particle further from the axis of rotation is greater

Question 40. Which of the following is a simple oscillatory motion?

1. Motion of the hand of a clock
2. Motion of the earth around the sun
3. Motion of a simple pendulum
4. Uniform circular motion

Answer: 3. Motion of a simple pendulum

Question 41. A man goes 4 m eastwards, then 4 m northwards and at the end, 3√2 m south- westwards. Displacement of the man is

1. 2√2m
2. √2m
3. 4m
4. 3√2m

Answer: 2. √2m

Question 42. During its journey, a car covers half of the total distance with a speed of 40 km/h and the rest half distance with a speed by 60 km/h. Average speed of the car is

1. 50 km/h
2. 46 km/h
3. 48 km/h
4. 52 km/h

Answer: 3. 48 km/h

Question 43. A ball is thrown vertically upward. If the rotation of the ball along its own axis is ignored then What type of motion will the ball follows?

1. Compound motion
2. Circular motion
3. Linear motion
4. Simple harmonic oscillation

Answer: 3. Linear motion

Question 44. A particle is revolving in a circular path of radius R. The displacement of the particle, when it completes one complete rotation, is

1. πR
2. 2πR
3. 0
4. 2R

Answer: 3. 0

Question 45. Distance travelled by a free falling body under the action of gravity in first 3s of motion is

1. g
2. \(\frac{3g}{2}\)
3. \(\frac{9g}{2}\)
4. \(\frac{9g}{4}\)

Answer: 3. \(\frac{9g}{2}\)

Question 46. Which of the following displacement-time dimensional graph represents one displacement of a particle?

Answer: 4.

Question 47. The velocity-time graph of a particle moving with uniform velocity is

1. A straight line parallel to the time axis
2. A straight line parallel to the velocity axis
3. A straight line passing through the origin
4. A curve line passing through the origin

Answer: 1. A straight line parallel to the time axis

Question 48. Velocity-time graph of a particle projected vertically upward is

Answer: 2.

Question 49. Initial and final velocity of a particle moving with constant acceleration are 30 cm/s and 40 cm/s respectively. Velocity of the particle at the midpoint of its way is

1. 35 m/s
2. 25√2 cm/s
3. 32 cm/s
4. 28√2 cm

Answer: 2. 25.2 cm/s

Question 50. A particle starts moving from its rest with constant acceleration and travel a certain distance in 100 s. Time required to travel first half of the distance is

1. 50s
2. 70s
3. 70.7s
4. 71s

Answer: 3. 70.7s

Question 51. A particle is projected vertically upward such that it attains maximum height ‘h’ and returns to ground. Here total distance traversed by the particle and total displacement of the particle are

1. h, 0
2. 0, 2h
3. 2h,0
4. 0, h

Answer: 3. 2h,0

Question 52. A body travels along a circular path of radius r and complete one rotation in time t: Here

1. average velocity = \(\frac{2 \pi r}{t}\)
2. total displacement = 2πr
3. total distance travel = 0
4. average speed = \(\frac{2 \pi r}{t}\)

Answer: 4. average speed = \(\frac{2 \pi r}{t}\)

Question 53. A piece of stone is dropped from the peak of a minar of height 20 m. What will be the speed of the stone when it hits the ground? (g = 10 m.s^-2)

1. 10 m.s^-1
2. 40 m.s^-1
3. 20 m.s^-1
4. 5 m.s^-1

Answer: 2. 40 m.s^-1

Question 54. In the velocity-time graph of a car is shown. Here the car moves with

1. Non-uniform velocity
2. Uniform acceleration
3. Uniform deceleration
4. Non-uniform deceleration

Answer: 3. Uniform deceleration

Question 55. Gradient of velocity-time graph of a particle is

1. Velocity
2. Acceleration
3. Distance traveled
4. Displacement

Answer: 2. Acceleration

Question 56. A boat first moves 12 m towards east and then 5 m towards west In a river. What is the total displacement of the boat?

1. 13m
2. 17m
3. 7m
4. 8m

Answer: 3. 7m

Question 57. SI unit of retardation is

1. -m.s^-2
2. cm.s^-2
3. m.s^-2
4. -cm.s^-2

Answer: 3. m.s^-2

Question 58. Rotation of the earth about it’s own axis is

1. Circular motion
2. Rotational motion
3. Translational motion
4. Compound motion

Answer: 3. Translational motion

Question 59. Motion of electron round a nucleus is

1. Circular motion
2. Rotational motion
3. Translation motion
4. Simple harmonic motion

Answer: 1. Circular motion

Question 60. Velocity of a particle moving in a straight line changes from 10 m.s^-1 to 20 m.s^-1 in time t s. If the particle travel 130 m distance in this time interval then value of t is 12

1. 1.8 s
2. 10 s
3. 12 s
4. 8.67 s

Answer: 4. 8.67 s

Question 61. Which of the following velocity-time graph represents the motion of a particle with non zero initial velocity and non uniform acceleration?

Answer: 1.

Question 62. Angle between displacement-time graph and time axis of two particles are 45° and 60° respectively. The Ratio of the velocity of two particles is

1. √3:1
2. 1:√3
3. 3:4
4. 1:1

Answer: 2. 1:3

Question 63. Average velocity of a particle is equal to its instantaneous velocity if the particle moves with

1. Constant acceleration
2. Constant speed
3. Gradually increasing speed
4. Constant velocity

Answer: 4. Constant velocity

Chapter 2 Topic A Rest Motion And Equation Of Motion Answer In Brief

Question 1. Imagining the earth as a point with respect to sun, what type of motion does the earth undergo around the sun?

Answer: In this case, rotation of the earth is a circular motion.

Question 2. What type of motion does a rotating body undergo around its own axis?

Answer: A rotating body undergoes a rotating motion around its own axis.

Question 3. Two bodies of masses 5 kg and 6 kg are falling from rest without any resistance. Which one has more acceleration?

Answer: Both the bodies fall with the same acceleration (acceleration due to gravity).

Question 4. Velocity-time (v-t) curve of a moving body is parallel to the time axis. What is the acceleration of the body?

Answer: Acceleration of the body is zero.

Question 5. Velocity-time (v-t) curves of two moving bodies form inclined angles of 30° and 60°. Which body has more acceleration?

Answer: The curve in respect of the body with inclined angle of 60° has comparatively more acceleration.

Question 6. Is it possible for the velocity-time (v-t) curve of a body to be perpendicular to the time-axis?

Answer: No, because in that case, the curve denotes different velocities of the body at a particular time which is impossible.

Question 7. What is the distance that a moving particle with initial velocity u and uniform deceleration a covers in time t?

Answer: Required distance, s = ut – \(\frac{1}{2}\) at².

Question 8. What is the velocity of a particle moving with initial velocity u and uniform acceleration a after traversing a distance s?

Answer: If v is the required velocity, then

v² = u² + 2at²

or, v = √u²+2as

Question 9. What is the change in acceleration of a body falling freely from a height without any resistance?

Answer: No, there is no change in acceleration of the body as this is acceleration due to gravity.

Question 10. What type of motion is the motion of wheel of a moving bicycle?

Answer: Motion of wheel of a moving bicycle is compound motion.

Question 11. What type of motion is the motion of an arm of a clock?

Answer: Motion of an arm of a clock is rotational motion or uniform circular motion.

Question 12. What type of motion is the motion of a freely falling body?

Answer: Motion of a freely falling body is translational motion.

Question 13. When do the passengers of two moving trains find each other to be stationary?

Answer: When two trains move in parallel with the same velocity, passengers find them to be stationary.

Question 14. What type of motion is the motion of a merry-go-round?

Answer: The motion of a merry-go-round is rotational motion.

Question 15. What is circular motion?

Answer: If a particle rotates around a fixed axis or a point in a circular path, then the motion of the particle is called circular motion.

Question 16. What is rectilinear oscillatory motion?

Answer: If a body moves repeatedly along the same straight line at equal intervals of time, then its motion is called rectilinear oscillatory motion.

Question 17. Can the distance traversed by a moving particle be zero?

Answer: No, the distance traversed by a moving particle cannot be zero.

Question 18. Can the displacement of a moving particle be zero?

Answer: If a particle starting from a particular point comes back to the same point, then the displacement of that particle is said to be zero.

Question 19. When displacement of a moving particle is zero?

Answer: Displacement of a moving is zero when its initial position and final position are the same.

Question 20. Can the ratio of distance traversed and displacement of a moving particle be less than 1?

Answer: No, the ratio of the distance travelled and the displacement of a moving particle is always 1 or more than 1.

Question 21. When is the ratio of the distance traversed and the displacement of a moving particle equal to 1?

Answer: The ratio of the distance traversed and the displacement of a moving particle is 1 when it moves in a straight line.

Question 22. When is the ratio of the distance traversed and the displacement of a moving particle more than 1?

Answer: The ratio of the distance traversed and the displacement of a moving particle is more than 1 when it moves in a curved path.

Question 23. What is the dimensional formula of speed?

Answer: Dimensional formula of speed is LT^-1.

Question 24. What type of physical quantity is speed?

Answer: Speed is a scalar quantity because it has magnitude, but no direction.

Question 25. A particle traverses the circumference of a circle of radius r and comes back to the position from where it started. What is its displacement?

Answer: Displacement of the particle is zero.

Question 26. What are the units of displacement in CGS system and SI?

Answer: Units of displacement in CGS system- centimetre (cm) and in SI-metre (m).

Question 27. A particle traverses the circumference of a circle of radius r and comes back to the position from where it started. What is the distance covered?

Answer: Distance covered = 2πr.

Question 28. Can the average speed of a moving particle be zero?

Answer: No, the average speed of a moving particle cannot be zero.

Question 29. Can the average velocity of a moving 38 What remains unchanged during a particle be zero?

Answer: Yes, the average velocity of a moving particle can be zero.

Question 30. What is negative acceleration called?

Answer: Negative acceleration is called deceleration or retardation.

Question 31. Displacement s of a particle in time t,s = 4t + 5t². If displacement of the particle is expressed in unit m and time in units, what is the initial velocity?

Answer: If we compare the equation s = 4t+5t² with the equation s = ut + 1/2 at², initial velocity is u = 4 m/s.

Question 32. What does the area made by a velocity-time graph with the time axis indicate?

Answer: It indicates the distance traversed.

Question 33. What is the nature of a displacement-time graph of a particle moving with uniform velocity?

Answer: The displacement-time graph of a particle moving with uniform velocity is a straight line passing through the origin.

Question 34. How do you define a reference body?

Answer: A reference body is a body with reference to which states of rest and motion of another body are measured.

Question 35. What is the direction of a particle moving in a circular path at a point?

Answer: The direction of a particle moving in a circular path at a point is along the tangent to the circle at that point.

Question 36. What does the odometer of a car indicate?

Answer: The odometer of a car indicates the instantaneous speed of the car.

Question 37. Can the average and the instantaneous speed of a moving particle be equal at any moment?

Answer: The average speed and the instantaneous speed of a particle moving with uniform speed are equal.

Question 38. What remains unchanged during a rotational motion?

Answer: The axis of rotation remains unchanged during rotational motion.

Question 39. 1 km/h = how many m/s?

Answer: 1 km/h = \(\frac{1000}{60 \times 60}\)m/s = \(\frac{5}{8}\) m/s

Question 40. How many times the words ‘per second’ come in the unit of acceleration?

Answer: The words ‘per second’ come twice in the unit of acceleration.

Question 41. What does the inclined angle of a displacement-time graph indicate?

Answer: Inclined angle of a displacement-time graph indicates the instantaneous velocity.

Question 42. What does the inclined angle of a velocity- time graph indicate?

Answer: Inclined angle of velocity-time graph indicates the instantaneous acceleration.

Question 43. What is the nature of the displacement-time graph of a particle moving with uniform acceleration?

Answer: The nature of the displacement-time graph of a particle moving with uniform acceleration is parabolic.

Question 44. What type of motion does a rotating body undergo around its own axis?

Answer: A rotating body under goes a rotational motion around its own axis.

Question 45. What is the direction of displacement?

Answer: Direction of displacement is considered from the initial position towards the final position of a body.

Question 46. s = 4t + 6t² is the displacement of a particle in time t. Considering displacement in m and time is s find acceleration of the particle.

Asnwer: Comparing s = ut + \(\frac{1}{2}\) at² and s = 4t + 6t² equations,

We get, = \(\frac{a}{2}\) = 6 or, a = 12

Therefore acceleration of the particle is 12 m.s^-2.

Question 47. What is the nature of velocity-time graph of a particle moving with constant acceleration and zero initial velocity?

Answer: Velocity-time graph of the particle moving with uniform acceleration and zero initial velocity is a straight line passing through the origin and inclined by an angle with the time axis.

Question 48. Initial velocity and uniform acceleration of a moving particle are A and B respectively. Find its velocity after time C.

Answer: Velocity of the particle after time C is = A + B X C. (Using v = u + at equation)

Question 49. What is the direction of acceleration of a particle moving in a circular path with uniform speed?

Answer: The direction of acceleration of a particle moving in a circular path with uniform speed is towards the ncentre of the circle (centripetal acceleration).

Question 50. Give an example where velocity and acceleration of a particle are directed opposite to each other.

Answer: Velocity and acceleration are directed opposite to each other for vertical motion of a body. Here gravitational acceleration directed opposite to the velocity of the particle.

Question 51. What are the units of acceleration in SI and CGS system?

Answer: Units of acceleration in CGS and SI system are cm-s-2 and m.s^-2 respectively.

Question 52. What are the units of velocity in SI and CGS system?

Answer: Units of velocity in CGS and SI system are cm.s^-2 and m.s^-2 respectively.

Question 53. Initial and final velocity of a particle moving with uniform acceleration are u and v respectively. what is its average velocity?

Answer: Average velocity of the particle is = \(\frac{u+v}{2}\)

Question 54. Draw velocity-time graph of a particle moving with uniform velocity.

Answer: Velocity-time graph of a particle moving with uniform speed is drawn. Here the graph is parallel to time axis.

Question 55. A train travels 10 km in 10 minute. What is the speed of the train in km/h unit?

Answer: 10 min = \(\frac{10}{60}h=\frac{1}{6}\)h

∴ speed of the train \(\frac{10}{\frac{1}{6}}\) km/h = 60 km/h

Chapter 2 Topic A Rest Motion And Equation Of Motion Fill In The Blanks

Question 1. Average _______ of a body may be zero but its average speed need not be zero.

Answer: Velocity

Question 2. Velocity-time (v-t) curve of a moving body with uniform acceleration starting from rest passes through the ______ of the graph.

Answer: Origin

Question 3. Starting from a particular place for a journey, if one comes back to the same place, total displacement is _______

Answer: Zero

Question 4. Speed is a ______ quantity.

Answer: Scalar

Question 5. If both the magnitude and the direction of the velocity of a particle remain unchanged with time, then velocity of that particle is called ________ velocity.

Answer: Uniform

Question 6. For a particle falling freely under gravity, constant quantity is ________

Answer: Acceleration

Question 7. Rest and motion for a particle are ______

Answer: Relative

Question 8. From a moving train, distant trees appear to be ________

Answer: Moving

Question 9. For pure rotation, the axis of rotation always remains _______

Answer: At rest

Question 10. Diurnal motion of the earth is an example of _______ motion.

Answer: Rotational

Question 11. Unit of ________ is used twice in the unit of acceleration.

Answer: Time

Question 12. The ______ of a body may not be zero even though its is zero.

Answer: Acceleration velocity

Question 13. Increase of velocity of a particle with time is called ________

Answer: Acceleration

Question 14. Velocity-time graph of a body moving from its rest with uniform acceleration is a straight line passing through the _______

Answer: Origin

Question 15. Mass of a body is its ______ property.

Answer: Intrinsic

Question 16. Retardation is also called _______ acceleration.

Answer: Negative

Question 17. The value of average velocity and average speed are same in _______ motion.

Answer: Translation

Question 18. Motion of simple pendulum is _______ in nature.

Answer: Periodic

Question 19. Area bounded by velocity-time graph and time axis divided by total time taken gives the physical quantity _______

Answer: Average velocity

Chapter 2 Topic A Rest Motion And Equation Of Motion State Whether True Or False

Question 1. If a body rotates around a fixed axis or a fixed point, its motion is called rotational motion.

Answer: True

Question 2. Rate of change of velocity of a particle with respect to time is called its acceleration.

Answer: True

Question 3. Average speed can be obtained by dividing the total displacement by the total time required to transverse the total distance.

Answer: False

Question 4. Dimensional formula of retardation is LT^-3

Answer: True

Question 5. Passengers of two trains running side by side with the same speed in the same direction, find them mutually at rest.

Answer: True

Question 6. Average velocity of a body will be zero when average speed of the body is zero.

Answer: False

Question 7. If speed is zero, velocity may not be zero.

Answer: False

Question 8. Speed may not be negative.

Answer: True

Question 9. v-t graph of a particle moving with uniform velocity is parallel to time axis.

Answer: True

Question 10. Free falling of a body is translational motion with constant acceleration.

Answer: True

Question 11. The annual motion of the centre of mass of the earth around the sun is rotational motion.

Answer: False

Question 12. Even if the instantaneous velocity of an object is zero, it may have acceleration.

Answer: True

Question 13. Particle having uniform speed must not have non-uniform velocity.

Answer: False

Question 14. Particle moving in a circular path may have uniform speed.

Answer: True

Question 15. Velocity and acceleration of an object must not have opposite directions.

Answer: False

Question 16. Gradient at any point in a v-t graph represents acceleration at that instant.

Answer: True

Chapter 2 Topic A Rest Motion And Equation Of Motion Numerical Examples

Useful Relations

If d be the distance covered by a particle in time t then speed, u = \(\frac{d}{t}\)

If distance covered by a particle in time \(t_1, t_2, t_3, \cdots, t_n \quad \text { are } \quad d_1, d_1, d_3, \cdots, d_n\) respectively, then

average speed, \(v_{a v g}=\frac{d_1+d_2+d_3+\cdots+d_n}{t_1+t_2+t_3+\cdots+t_n}\)

If s be the displacement of a moving particle in time t then velocity, v = \(\frac{s}{t}\)

If the initial velocity of a particle is u and the final velocity of the particle after time t will be v, then acceleration, a = \(\frac{v-u}{t}\)

If initial velocity of a particle = u, final velocity of the particle = v, acceleration = a, time taken = t, displacement = s then,

1. \(v=u+a t\)
2. \(s=u t+\frac{1}{2} a t^2\)
3. \(v^2=u^2+2 a s\)

If the initial velocity, u = 0,

1. v = at
2. s = \(\frac{1}{2}\)at²
3. v = √2as

If a be the retardation,

1. v = u – at
2. s = ut – \(\frac{1}{2}\)at²
3. v² = u²-2as

Question 1. Two individuals start their journey from C and reach B. One goes through the path CDAB and the other through the path CDB. ABCD is a square whose length of each side is 2 m. What are the displace ments of C and B? How much distance does each one cover?

Answer:

As both of them start from initial position C and reach final position B, both have the displacement, CB = 2 m.

Distance covered by the first individual

= CD + DA + AB = 2 + 2 + 2 = 6m

Distance covered by the second individual

= CD + DB = 2 + 2√2 = 2(1+ √2) m

Concepts Related to Speed, Velocity, and Acceleration for Class 9 Solutions

Question 2. A moving particle goes from one end to the other end of a semicircular path of radius 7 cm. Calculate the total distance cov- ered and the magnitude of displacement.

Answer:

Radius, r = 7 cm

∴ Distance Covered

= Circumference of the semicircle

= πr = \(\frac{22}{7}\) x 7 = 22 cm

∴ displacement 2r = 2 x 7 = 14 cm

Question 3. A particle moves from point A to point B along the circumference of a circle of radius 10√2 cm such that an angle of 90° is formed at the centre. What displacement of the particle?

Answer:

Suppose, centre of circle is at point O.

Radius of the circle = 10√2 cm

According to AOB is a right-angled triangle.

∴ \(A B^2=O A^2+O B^2=r^2+r^2=2 r^2\)

or, AB = \(\sqrt{2} r=\sqrt{2} \times 10 \sqrt{2}=20 \mathrm{~cm}\)

∴ Displacement of the particle is 20 cm.

Question 4. A wheel makes half a rotation on a plane surface. If radius of this wheel is 21 cm, what is the displacement of the point touching the ground?

Answer:

Initial and final positions of the wheel are shown respectively.

Point A touching the ground has moved to the position A₁ after making half a rotation.

At the final position, B₁ is the point touching the ground.

Radius of the wheel, r = 21 cm

∴ \(A B_1=\pi r=\frac{22}{7} \times 21=66 \mathrm{~cm}\)

and \(A_1 B_1=2 r=2 \times 21=42 \mathrm{~cm}\)

So, \(A A_1=\sqrt{\left(A B_1\right)^2+\left(A_1 B_1\right)^2}\)

= \(\sqrt{(66)^2+(42)^2}=\sqrt{6120}=78.23 \mathrm{~cm}\)

∴ Displacement of the point touching the ground is 78.23 cm.

Question 5. During the first half of its journey, a car moves with a speed u and it makes the remaining half of the journey with a speed v. What is the average speed of this car?

Answer:

Suppose, distance to be covered = 2s.

Time taken to cover the first half of the journey, \(t_1=\frac{s}{u}\)

Time taken to cover the second half of the journey, \(t_2=\frac{s}{v}\)

∴ Average Speed,

\(v_a=\frac{2 s}{t_1+t_2}=\frac{2 s}{\frac{s}{u}+\frac{s}{v}}=\frac{2}{\frac{1}{u}+\frac{1}{v}}=\frac{2}{\frac{u+v}{u v}}=\frac{2 u v}{u+v}\)

Question 6. A particle moves along the circumference of a circle of radius 21 cm from point A to point B with uniform speed and forms an angle of 60° at the centre. What is the speed of the particle?

Answer:

Circumference of a circle of radius 21 cm

= \(2 \pi r=2 \times \frac{22}{7} \times 21=132 \mathrm{~cm}\)

After completion of one round along the circumference, the particle forms an angle of 360° at the centre with respect to the line joining the initial position of the particle (A) and the centre of circle (O).

But as the particle forms an angle of 60° at the centre, distance traversed by it is given by

s = \(2 \pi r \times \frac{60^{\circ}}{360^{\circ}}=132 \times \frac{1}{6}=22 \mathrm{~cm}\)

Now, Time Spent, t = 2 s

∴ Speed of the particle, v = \(\frac{s}{t}=\frac{22}{2}\) = 11 cm/s

Question 7. A train travels the first 30 km of its total journey of 100 km at the uniform speed of 30 km/h. What is its speed for the remaining 70 km so that the average speed of the total journey is 40 km/h?

Answer:

Train travels first s₁ = 30 km of the journey at the uniform speed of v₁ = 30 km/h.

Let the train travel remaining s₂ = 70 km distance at uniform speed v₂ so that its average speed for the total journey, v_a becomes 40 km/h.

If time t₁ and t₂ are taken to cover the first 30 km and the last 70 km respectively, then

\(v_a=\frac{s_1+s_2}{t_1+t_2} \quad \text { or, } v_a=\frac{s_1+s_2}{\frac{s_1}{v_1}+\frac{s_2}{v_2}}\)

 

or, \(40=\frac{30+70}{\frac{30}{30}+\frac{70}{v_2}} \quad \text { or, } 1+\frac{70}{v_2}=\frac{100}{40}\)

or, \(\frac{70}{v_2}=2.5-1\)

∴ \(v_2=\frac{70}{1.5}=46.67 \mathrm{~km} / \mathrm{h}\)

Question 8. A train goes from Dum Dum to Naihati with uniform speed u and comes back from Naihati to Dum Dum with uniform speed v. What is the average speed of this train?

Answer:

Let the distance between Dum Dum and Naihati is s.

∴ Time taken to go from Dum Dum to Naihati with uniform speed u, \(t_1=\frac{s}{u}\)

During the return journey, time taken to cover this same path with uniform speed v, \(t_2=\frac{s}{v}\)

∴ Average speed of the train

= \(\frac{\text { total distance }}{\text { total time }}=\frac{s+s}{t_1+t_2}=\frac{2 s}{\frac{s}{u}+\frac{s}{v}}\)

= \(2 s \times \frac{u v}{s u+s v}=\frac{2 s \times u v}{s(v+u)}=\frac{2 u v}{u+v}\)

Question 9. A car picks up a velocity of 45 km/h after 10s of starting. Calculate the acceleration of the car.

Answer:

Velocity of the car after t = 10 s of starting,

v = \(45 \mathrm{~km} / \mathrm{h}=\frac{45 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=12.5 \mathrm{~m} / \mathrm{s}\)

Suppose, acceleration of the car = a

∴ \(v=a t \text { or, } a=\frac{v}{t} \text { or, } a=\frac{12.5}{10}=1.25 \mathrm{~m} / \mathrm{s}^2\)

Example 10. A train stops after 10s of application of brakes. If deceleration of the train is 3 m/s², what is its velocity at the time of application of the brakes?

Answer:

Suppose, velocity of the train at the time of application of brakes is u.

Deceleration of the train, a = 3 m/s²

The train stops after time t = 10 s

From the equation v=u-at, we get

0 = u – at [v=0]

or, u = at = 3 x 10 = 30 m/s

Question 11. A train running with a velocity of 10 m/s attains an acceleration of 2 m/s². What is the distance covered by it in 10s?

Answer:

Initial velocity (u) = 10 m/s and acceleration (a) = 2 m/s².

A train traverses a distnce s in t = 10 s.

∴ \(s=u t+\frac{1}{2} a t^2=10 \times 10+\frac{1}{2} \times 2 \times 10^2\)

= 100 + 100 = 200 m

Question 12. A car travels y km in the first x min and x km in the next y min. What is the average velocity of this car?

Answer:

Total distance traversed, s = (y + x) km

Total time required, t = (x + y) min

= \(\frac{x+y}{60} \mathrm{~h}\)

∴ Average velocity of this car,

\(v_a=\frac{s}{t}=\frac{x+y}{\frac{x+y}{60}}=60 \mathrm{~km} / \mathrm{h}\)

Question 13. A train was running with a velocity of 72 km/h. It took a time of 40 s to stop after the brakes were applied. Calculate the distance traversed by the train and its deceleration after the application of brakes.

Answer:

Initial velocity of the train,

\(u=72 \mathrm{~km} / \mathrm{h}=\frac{72 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}\)

Velocity of train, v = 0 after t = 40 s

Let deceleration of the train = a.

From the equation v = u – at, we get

0 = u – at or, at = u

or, \(a=\frac{u}{t}=\frac{20}{40}=0.5 \mathrm{~m} / \mathrm{s}^2\)

If the train traverses a distance s before it stops, v² = u²-2as gives

\(0=u^2-2 a s \quad \text { or, } 2 a s=u^2\)

∴ \(s=\frac{u^2}{2 a}=\frac{20^2}{2 \times 0.5}=400 \mathrm{~m}\)

Question 14. A particle starts from rest and covers a distance of 200 m with uniform accelera- tion. What percentage of total time is required to cover half the distance?

Answer:

Suppose, acceleration of the particle = a and t time is taken to cover a distance of s = 200 m.

∴ \(s=\frac{1}{2} a t^2\)

If half the disnce \(s_1=\frac{s}{2}=100 \mathrm{~m}\) is traversed in time, t₁ then

∴ \(s_1=\frac{1}{2} a t_1^2 \quad \text { or, } \frac{s}{2}=\frac{1}{2} a t_1^2\)

or, \(\frac{1}{2} \times \frac{1}{2} a t^2=\frac{1}{2} a t_1^2\) [from equation (1)]

or, \(t_1^2=\frac{t_2}{2}\)

or, \(t_1=\frac{t}{\sqrt{2}}=\frac{\sqrt{2}}{2} t=\frac{1.414}{2} t=0.707 t\)

∴ \(t_1=70.7 \% of t\)

Question 15. A bullet loses half of its velocity after entering 6 cm in a wooden block. How much more distance does it penetrate to come to a halt?

Answer:

Suppose the deceleration of the bullet in the wooden block a and its initial velocity = u.

If the velocity of the bullet is \(v_1=\frac{u}{2}\) after it traverses a distance s₁ = 6 cm.

∴ \(v_1^2=u^2-2 a \cdot s_1 \quad \text { or, } \frac{u^2}{4}=u^2-2 \cdot a \cdot 6\)

or, \(12 a=u^2-\frac{u^2}{4}=\frac{3 u^2}{4} \quad \text { or, } a=\frac{u^2}{16}\)

Its final velocity is V = 0 as it comes to rest after traversing a further distance.

So, \(v^2=v_1^2-2 a s\)

or, \(0=\frac{u^2}{4}-2 \cdot a s \quad or, 2 a s=\frac{u^2}{4}\)

or, \(2 \times \frac{u^2}{16} \times s=\frac{u^2}{4} \quad or, \frac{s}{8}=\frac{1}{4}\)

∴ \(s=\frac{8}{4}=2 \mathrm{~cm}\)

Question 16. A bullet running with a velocity enters a wooden block. After moving x distance in it, its velocity becomes v and after moving a further distance y, it comes to rest. Show that \(\frac{y}{y}=\sqrt{\frac{y}{x+y}}\).

Answer:

Suppose the deceleration of the bullet in the wooden block = a and it stops after traversing a total distance of (x + y).

∴ 0 = u² – 2a(x + y)

or, u2 = 2a(x+y) ……(1)

Again, when its velocity is v, it stops after traversing an additional distance y.

∴ 0 = v² – 2ay or, v² = 2ay…..(2)

Dividing equation (1) by (2), we get

\(\frac{v^2}{u^2}=\frac{y}{x+y}\)

∴ \(\frac{v}{u}=\sqrt{\frac{y}{x+y}}\)

Study Guide for Class 9 Force and Motion Questions

Question 17. Velocity-time (v-t) graph of a particle is shown in the following diagram.

1. What is the highest velocity of the particle?
2. What is the acceleration of the particle?
3. How much distance does it cover with uniform acceleration?
4. How much distance does it cover with uniform velocity?
5. What is the deceleration of the particle?
6. How much distance does it cover with uniform deceleration?
7. How much total distance does the particle cover?

Answer:

It is seen from the graph that the particle moves with uniform acceleration for the first 2 s to attain a velocity of 10 m/s. Then it travels with a uniform velocity of 10 m/s for 4s (from 2s to 6s).

It travels with uniform deceleration to come to rest in the last 2s(6 s to 8 s).

1. Highest velocity of the particle = AD = 10 m/s

2. Acceleration during the first 2 s

= \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time }}\)

= \(\frac{(10-0) \mathrm{m} / \mathrm{s}}{2 \mathrm{~s}}=5 \mathrm{~m} / \mathrm{s}^2\)

3. Distance traversed by the particle with uniform acceleration,

s₁ = area of AOAD = \(\frac{1}{2}\) × OD X AD

= \(\frac{1}{2}\) x 2s x 10 m/s = 10 m

4. Distance traversed by the particle with uniform velocity,

s₂ = area of rectangle ABED

=AD X DE = 10 m/s x (6-2)s

= 40m

5. Deceleration of the particle in the last 2s

= \(\frac{\text { initial velocity }- \text { final velocity }}{\text { time }}\)

= \(\frac{(10-0) \mathrm{m} / \mathrm{s}}{2 \mathrm{~s}}=5 \mathrm{~m} / \mathrm{s}^2\)

6. Distance traversed by the particle with uniform deceleration,

\(s_3=\text { area of } \triangle B E C=\frac{1}{2} \times E C \times B E\)

= \(\frac{1}{2} \times(8-6) \times 10=\frac{1}{2} \times 2 \times 10=10 \mathrm{~m}\)

7. Distance traversed by the particle in total 8 seconds,

s = s₁ + s₂ + s₃ = 10 + 40 + 10 = 60 m

Question 18. Velocity-time (v-t) graph of a particle is shown in the following diagram. Determine the average velocity of the particle.

Answer:

Distance traversed by the particle,

s = \(\text { area of } \triangle O A B=\frac{1}{2} \times O B \times A C\)

= \(\frac{1}{2} \times 10 \times 10=50 \mathrm{~m}\)

∴ Average velocity of the particle,

\(v_a=\frac{\text { total displacement }}{\text { total time }}=\frac{50}{10}\)

= 5 m/s

Question 19. A particle moves in a straight line with uniform acceleration. Velocity of the particle at different times is given In the following chart:

Draw The Velocity-Time Graph Of The Particle.

Determine From The Graph:

1. Velocity after 3 s,
2. Acceleration of the particle and
3. Distance traversed during the last 6 s.

Answer:

Velocity-time (v-t) graph of the particle is

1. From the graph, velocity of the particle after 3s, v₁ = BC = 6 m/s

2. Acceleration of the particle, \(a=\frac{B C}{O B}=\frac{6}{3}=2 \mathrm{~m} / \mathrm{s}^2\)

3. Distance traversed during the last 6 s time,

s = area of trapezium DEFG

= \(\frac{1}{2}\)(DE + FG) X DG

= \(\frac{1}{2}\)(8 + 20) X (10 – 4)

= \(\frac{1}{2}\) X 28 X 6 = 84 m

Question 20. A particle starts from rest and attains a velocity of 40 cm/s after traversing for 10s with uniform acceleration. Next, the particle moves for 10s with uniform velocity. Finally, the particle comes to rest while moving for 10s with uniform deceleration. Draw the velocity-time graph for the partide. Calculate the total distance traversed from the graph.

Answer:

Velocity-time (v-t) graph of the particle is shown where OA is the graphical representation of time 0-10 s during uniform acceleration state.

Further, AB and BC are graphical representations of the uniform velocity period during 10s-20s and uniform deceleration period during 20s – 30s, respectively.

∴ Total distance traversed by the particle,

s = area of trapezium OABC

= \(\frac{1}{2}\)(OC + AB) X AE = \(\frac{1}{2}\)(30 + 10) X 40

= \(\frac{1}{2}\) x 40 x 40 = 800 cm = 8 m

Question 21. A ball is thrown vertically upward with a velocity 50 m/s. Calculate the time taken by the ball to return at the point of ejection. [g = 10m-s-2].

Answer:

Initial velocity of the ball is u = 50 m • s^-1.

Let, t be the required time.

∴ Total height covered by the ball in time t is,

h = 0

From equation h = ut-1/2 gt² we get,

0 =(50.t) – (1/2 .10.t¹)

or, t(50 – 5t) = 0

∴ t = 0 or, 50 – 5t = 0 or, t = 50/5 = 10

(t = 0 means initial position of the ball)

Therefore, the required time is 10 s.

Question 22. A car starts moving with an initial velocity and uniform acceleration. It traversed 81 m distance in 5 s. After that it
travels 72 m with uniform velocity in next 4s. Determine the initial velocity and acceleration of the car.

Answer:

Let, the initial velocity and the acceleration of the car be u m/s and a m • s^-2.

Final velocity of the car after 5 s is v m/s .

The car travels 72 m distance in 4 s with uniform velocity.

∴ v = 72/4  = 18 m/s

Now from equation v = u + at we get,

18 = u + a x 5

∴ u = 18 – 5a

Here, t = 5 s, u = 18 m/s

Again from equation v² = u² + 2as we get,

18² = (18- 5a)² + 2 x a • 81 [s = 81 m]

or, 18² = 18² – 2 • 18 • 5a + 25a²+ 162a

or, 25a² -(180-162)a = 0

or, a[25a- 18] = 0

∴ a = 0

or, a = 18/25 = 0.72 [Here, car moves with uniform acceleration.  ∴ a ≠ 0]

∴ Acceleration of the car is 0.72 m • s^-2 and initial velocity,

u = 18 – 5a = 18 – 5 x 0.72

= 18- 3.6 = 14.4 m • s^-2