One-Dimensional Motion
One-Dimensional Motion Introduction: Mechanics is the branch of science which deals with the motion of bodies and the effect of force on stationary or moving bodies.
The Branches Of MechanicsAre As Follows:
- Dynamics: A branch of mechanics in which the motion of a body and the effect of force on the motion of the body are discussed.
- Kinematics: It is restricted to the study of motion and not the causes of motion.
- Kinetics: It incorporates the study and analysis of the motion of a body together with the causes of motion. Also, the mass of a body and the effect of force on the mass are studied here.
- Statics: A branch of mechanics which deals with the equilibrium of a body under the action of a number of forces and studies the conditions of equilibrium.
One-Dimensional, Two-Dimensional And Three-Dimensional Motions: We will be discussing only one-dimensional motion in this chapter. The motion which is confined to a straight line is called one-dimensional motion or rectilinear motion.
- This type of motion can be explained by one-dimensional, two-dimensional or three-dimensional coordinates. The motion of a car moving in a straight line or the motion of a body under gravity are examples of one-dimensional motion.
- The motion which is confined to a plane is called two-dimensional motion or planar motion is called two- dimensional motion can only be described by two-dimensional or three-dimensional coordinates. The motion of a planet around the sun, a body revolving in a circle, a billiard ball moving over the billiard table etc. are examples of two-dimensional motion.
- The motion which is not confined to a plane is called three-dimensional motion. This type of motion can only be explained by three-dimensional coordinates, flic examples of three-dimensional motion are the spiral motion of a particle or the motion of an aeroplane
Read and Learn More: Class 11 Physics Notes
Speed And Velocity Numerical Examples
Example 1. A particle moves in a circular path of radius 7 cm. It covers
- Half of the circle in 4 s and
- One complete round in 10 s. In each case find the average speed and average velocity.
Solution:
The circumference of the circle = 2πr = 2 x \(\frac{22}{7}\) x 7 = 44cm
1. The distance travelled by the particle = half of the circumference = \(\frac{44}{2}\) = 22 cm and time taken = 4 s.
∴ Average speed, v = \(\frac{22}{4}\) = 5.5 cm · s-1.
Displacement = diameter of the circle = 2 x 7 = 14 cm and time taken = 4s.
Hence average velocity, \(\vec{v}\) = \(\frac{14}{4}\) = 3.5 cm · s-1 along AB.
2. The distance travelled by the particle = circumference of the circle = 44 cm and time taken = 10 s.
∴ Average speed = = 4.4 cm · s-1.
Displacement in this case is 0 (as initial and final positions are the same.)
Hence, average velocity = \(\frac{\text { displacement }}{\text { time }}\)
Example 2. An aeroplane travels 2000 km to the west It then turns north and moves 2000 km more. Finally, it follows the shortest path to return to its starting point If the speed of the plane is 200 km · h-1, find its average velocity for the total journey.
Solution:
Since initial and final positions are the same, displacement is zero.
∴ Average velocity = \(\frac{\text { displacement }}{\text { total time }}=\frac{0}{\text { total time }}=0 \text {. }\)
Example 3. Find the speed of the tip of a 3 cm long second’s hand in a clock.
Solution:
The tip of the second’s hand describes an angle of 360° in 60 seconds when it completes the total circular path once.
Hence, distance travelled = circumference of the circle = 2π x 3 = 6π cm; time = 60 s.
∴ Speed = \(=\frac{6 \pi}{60}=\frac{\pi}{10}=0.314 \mathrm{~cm} \cdot \mathrm{s}^{-1} .\)
Example 4. A train travels from station A to station B at a constant speed of 40 km · h-1 and returns from B to A at 60 km · h-1. Find the average speed and average velocity of the train.
Solution:
Let the distance between stations A and B be x km.
Time taken by the train to move from A to B = \(\frac{x}{40}\)h and time taken to move from B to A = \(\frac{x}{60}\)h.
Total distance travelled = 2x km.
∴ Average speed = \(\frac{2 x}{\left(\frac{x}{40}+\frac{x}{60}\right)}=\frac{2 \times 40 \times 60}{100}=48 \mathrm{~km} \cdot \mathrm{h}^{-1}\)
Initial and final positions are the same so the total displacement becomes zero and hence average velocity is zero.
Example 5. The motion of a particle, along the x-axis, follows the relation x = 8t – 3t². Here x and t are expressed in metre and second respectively. Find
- The average velocity of the particle in time interval 0 to 1 s and
- Its instantaneous velocity at t = 1s.
Solution:
1. Let at t = 0, x = x1 and at t = 1, x = x2.
∴ x1 = 8 x 0 -3 x 0² = 0 and x2 = 8 x 1 -3 x 1² = 5m
∴ Displacement = x2 – x1 =5 m and time taken is 1 s
∴ Average velocity = \(\frac{5 m}{1 s}\) = 5 m · s-1.
2. Here instantaneous velocity,
v = \(\frac{d x}{d t}=\frac{d}{d t}\left(8 t-3 t^2\right)=8-2 \times 3 t=8-6 t\)
At t = \(1 \mathrm{~s}, \frac{d x}{d t}=8-6 \times 1=2 \mathrm{~m} \cdot \mathrm{s}^{-1}\).
WBBSE Class 11 One Dimensional Motion Notes
Acceleration: When the velocity of a particle increases with time, the particle is said to be accelerating. So, in case of acceleration, the final velocity is greater than the initial velocity.
Acceleration Definition: The rate of change of velocity with time is called acceleration.
Thus, acceleration (a) = \(\frac{\text { change in velocity }}{\text { time }}\)
= \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time }}\)
Acceleration Example: A train at rest starts from a station and speeds up. In this case, we can say that the train is moving with acceleration.
Sometimes, acceleration is represented by the symbol ‘f’.
Nature Of Acceleration: Acceleration is related to the change in velocity of a body. So, like velocity, acceleration is also a vector quantity. It has to be specified by its magnitude and direction. However, to specify the acceleration vector, we have to use vector algebra to determine the change in the velocity.
This shows that velocity and change in velocity may have different directions, in general. Thus, the direction of the acceleration may or may not be the same as that of velocity.
Units And Dimension: Unit of acceleration = \(\frac{\text { unit of velocity }}{\text { unit of time }} \text {. }\)
CGS System: cm · s-2
SI: m · s-2
Dimension of acceleration = \(\frac{\text { dimension of velocity }}{\text { dimension of time }}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}}=\mathrm{LT}^{-2}\)
Motion With Uniform And Non-Uniform Acceleration: Uniform acceleration corresponds to a motion in which the velocity of a body changes equally in equal intervals of time.
If a particle moves with uniform acceleration, then its acceleration remains the same, both in magnitude and direction, at each point on its path. When a body falls freely from a height under gravity, its velocity increases. But its acceleration is uniform on or near the surface of the earth and is known as the acceleration due to gravity.
Non-uniform Acceleration corresponds to a motion in which the velocity of a body does not change equally in equal intervals of time.
The motion of an oscillating pendulum is an example of motion with non-uniform acceleration. The acceleration of the bob becomes maximum at its maximum displaced position and becomes zero at its mean position.
Average Acceleration: The acceleration of a particle may not always be uniform. Generally, we can find out the average acceleration using the following relation:
(a) = \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time taken }}\)
= \(\frac{\text { change in velocity }}{\text { time }}\)
Instantaneous Acceleration: The acceleration of a particle at any moment is called its instantaneous acceleration.
Instantaneous Acceleration Definition: The instantaneous acceleration of a particle at a given point is the limiting value of the rate of change in velocity with respect to time when the time interval tends to zero.
According to differential calculus, instantaneous acceleration, a = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}=\frac{d}{d t}\left(\frac{d s}{d t}\right)=\frac{d^2 s}{d t^2}\)
Deceleration Or Retardation: when the velocity of a particle decreases with time, the particle is in a state of deceleration or retardation. Here, the final velocity is less than the initial velocity.
Deceleration Or Retardation Definition: The rate of decrease of velocity with time is called deceleration or retardation.
Deceleration Or Retardation Example: When a train approaches a station, it slows down and finally stops. During this period the train decelerates to come to a halt.
Deceleration is a special case of acceleration where the final velocity is less than the initial velocity. Thus, deceleration is essentially a negative acceleration. It is not a different physical quantity.
Let us consider a particle moving with a velocity of 10 cm · s-1. It then slows down to 4 cm · s-1 in 3 s. Its acceleration is given by,
a = \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time interval }}\)
= \(\frac{4-10}{3}=-2 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)
This example clearly shows that deceleration is described mathematically as negative acceleration.
Distinction Between Acceleration And Retardation
Acceleration Due To Gravity: The earth attracts other bodies towards itself because of its gravity. A body released near the earth’s surface falls freely under the action of the force due to gravity. Whenever a body is allowed to fall freely, it undergoes an acceleration directed towards the earth. This is referred to as the acceleration due to gravity and is denoted by g. Its characteristics are:
- It is always directed towards the centre of the earth, i.e., vertically downwards.
- Its value is the greatest on the earth’s surface. It decreases slightly as we go away from the surface in such a way that at an altitude of 3.2 km, its value decreases only by 0.1%. Thus, for practical purposes, we may consider its value to be constant and equal to its value on the earth’s surface.
- It is independent of the characteristics of an object, such as mass, density or shape.
Bodies that move vertically downwards undergo constant acceleration, while bodies moving vertically upwards undergo constant retardation. In this discussion, we have neglected the small variations in g at different locations and the air resistance faced by freely falling bodies.
The value of g is usually taken as, g = 980 cm · s-2 = 9.8 m · s-2
Understanding Displacement in One Dimensional Motion
Motion In A Straight Line
When a body moves without changing its direction, the motion is naturally along a straight line. It is also known as rectilinear motion.
Vertical fall of a body under gravity, the motion of a car on a straight road, etc. are examples of such motion. The discussions henceforth in this chapter will be restricted to rectilinear motions only.
Representation Of The Physical Quantities Of Motion: For the rectilinear motion of a particle, the straight line of motion itself may be chosen as an axis (say, the x-axis) and a point O on it as the origin. Obviously, the motion is one-dimensional.
- There is a basic difference in the representations of the scalar and the vector quantities related to rectilinear motion. The scalar quantities like distance travelled and speed have magnitudes only, and are always expressed by positive numbers. However, two directions, exactly opposite to each other, exist for the motion in a straight line.
- So a vector quantity is expressed by a positive number for one direction and by a negative number for the exactly opposite direction. The direction towards the right may be taken as positive; then the direction towards the left naturally becomes negative. As an example, we may consider the vertical motion of a particle under gravity.
- For a downward motion, we may take each of displacement, velocity and acceleration to be positive. Then, for an upward motion, displacement and velocity would be negative; but acceleration would still remain positive since the acceleration due to gravity (g) is always directed downwards.
- It is important to note that simple algebraic operations are sufficient for calculations involving any quantity, a scalar or a vector, in a rectilinear motion. It means that vector algebra is not at all necessary even for the calculations involving vector quantities of a rectilinear motion.
For motion along a straight line,
- Distance travelled by a particle = magnitude of its displacement,
- Speed of the particle at any point = magnitude of its velocity at that point
- Displacement, velocity and acceleration vectors are along the same straight line.
Rectilinear Motion with Uniform Velocity: If a particle moves with a uniform velocity, its acceleration is zero. Let v be the uniform velocity of a particle and s be its displacement in time t. Therefore, according to the definition of uniform velocity, the particle moves a distance v x 1 in Is, v x 2 in2s, etc.
∴ In t s its displacement is v x t.
∴ s = vt…..(1)
i.e., displacement = uniform velocity x time .
Motion In A Straight Line Numerical Example
Example 1. A person travels half of a distance at an average velocity of 24 km · h-1. At what average velocity should he move to cover the second half of the path so that his average velocity for the total path becomes 32 km · h-1?
Solution:
Let the total length of the path = 2s km.
∴ Time required to cover the first half of the path = \(\frac{s}{24}\) h.
If the man travels the total path with an average velocity of 32 km · h-1, then the total time taken by him = \(\frac{2 s}{32}\) = \(\frac{s}{16}\) h.
∴ Time required to cover the second half of the path = \(\frac{s}{16}\) – \(\frac{s}{24}\) = \(\frac{s}{48}\) h
∴ Average velocity of second half = \(\frac{\text { distance }}{\text { time }}=\frac{s}{\frac{s}{48}}=48 \mathrm{~km} \cdot \mathrm{h}^{-1}\)
Rectilinear Motion With Uniform Acceleration
Rectilinear Motion With Uniform Acceleration: For a particle in motion, let
u = initial velocity
v = final velocity after a time t and
corresponding uniform acceleration = a.
Let the displacement = s in that time.
Then the above variables obey the following equations:
- v = u+at,
- s= ut+1/2 at²
- v² = u² + 2as
- st = M + \(\frac{1}{2}\) a(2t – 1) [where st = the displacement in the t th second]
Rectilinear Motion With Uniform Acceleration Derivations:
1. v= u+at
Let the initial velocity of a particle be u and its final velocity after time t be v.
∴ In time t, change in velocity = v – u
∴ Rate of change of velocity with time = \(\frac{v – u}{t}\) = a, by definition.
Hence, at = v – u or, v = u + at….(1)
For a particle starting from rest, u = 0 and thus, v = at ….(2)
In case of retardation, the relationship becomes v = u – at …(3)
2. s = ut + \(\frac{1}{2}\) at²
If a particle with initial velocity u and uniform acceleration a, attains final velocity v after time t, the average velocity, considering the two endpoints, is \(\frac{u + v}{2}\).
The acceleration a implies that the velocity increases by an after every second. Hence, the velocity 1 s after the start of motion = u + a and 1 s before the end of motion = v – a
Velocity and Speed in One Dimensional Motion
∴ Average velocity = \(\frac{u+a+v-a}{2}=\frac{u+v}{2}\)
Hence, the particle can be considered to have travelled a distance s with the average velocity \(\frac{(u+v)}{2}\) in time t.
Hence, displacement, s = \(\frac{u+v}{2} \times t=\frac{u+(u+a t)}{2} \times t\) (because v=u+a t)
or, \(s=\frac{2 u t}{2}+\frac{1}{2} a t^2\) or, \(s=u t+\frac{1}{2} a t^2\)…..(4)
For a particle starting from rest, u=0
So, \(s=\frac{1}{2} a t^2\)…..(5)
For a retarding particle, s = \(u t-\frac{1}{2} a t^2\)…..(6)
3. \(v^2=u^2+2 a s\)
From equation (1) we have, v=u+a t
or, \(v^2=(u+a t)^2=u^2+2 u a t+a^2 t^2\)
= \(u^2+2 a\left(u t+\frac{1}{2} a t^2\right)\)
or, \(v^2=u^2+2\) as [using equation (4)]…….(7)
Hence, for a particle starting from rest \(\nu^2=2 a s\)……(8)
and in case of retardation \(v^2=u^2-2 a s\)……(9)
4. \(s_t=u+\frac{1}{2} a(2 t-1)\)
Displacement in t seconds, s = \(u t+\frac{1}{2} a t^2\), from equation (4).
Displacement in (t-1) seconds, \(s^{\prime}=u(t-1)+\frac{1}{2} a(t-1)^2\)
Hence, the displacement in the t th second, \(s_t=s-s^{\prime}\)
or, \(s_t=u t+\frac{1}{2} a t^2-\left\{u(t-1)+\frac{1}{2}(t-1)^2 a\right\}\)
= \(u t+\frac{1}{2} a t^2-u t+u-\frac{1}{2} a t^2+\frac{1}{2} \cdot 2 t a-\frac{a}{2}\)
= \(u+a t-\frac{a}{2}=u+\frac{1}{2} a(2 t-1)\)…….(10)
For a particle starting from rest \(s_t=\frac{1}{2} a(2 t-1)\)……(11)
and in case of retardation \(s_t=u-\frac{1}{2} a(2 t-1)\)…..(12)
Rectilinear Motion With Uniform Acceleration Numerical Examples
Example 1. A velocity of 60 km · h-1 of a train is reduced by the application of brakes. A retardation of 40 cm · s-2 is produced. After how much time will the train stop? What will be the velocity of the train after 20 seconds?
Solution:
Given, \(u=60 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{60 \times 1000}{60 \times 60}=\frac{50}{3} \mathrm{~m} \cdot \mathrm{s}^{-1}\)
a = 40 cm · s-2 = 0.4 m · s-2 and v = 0
Hence, from the relation v = u- at, we get \(0=\frac{50}{3}-0.4 \times t \quad \text { or, } t=\frac{50}{3 \times 0.4}=41.7 \mathrm{~s}\)
∴ The train will stop after 41.7.
The velocity after 20 s, v = \(\frac{50}{3}-0.4 \times 20\)
= \(\frac{26}{3} \mathrm{~m} \cdot \mathrm{s}^{-1}=8.7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
= \(\frac{8.7 \times 60 \times 60}{1000} \mathrm{~km} \cdot \mathrm{h}^{-1}\)
= \(31.3 \mathrm{~km} \cdot \mathrm{h}^{-1}\)
Example 2. A body covers 200 cm in the first 2 s of motion and 220 cm in the next 4 s. Calculate the velocity 7 s after the start.
Solution:
We know, s = ut + \(\frac{1}{2}\) at²
∴ Putting t = 2 s
200 = u x 2 + \(\frac{1}{2}\) a x 4 = 2u + 2a
or, u + a = 100….(1)
The displacement in (4 + 2) or 6 s = 200 + 220 = 420 cm
∴ 420 = u x 6 + \(\frac{1}{2}\) a x 36 = 6u + 18a
or, u + 3a = 70…..(2)
From equations (1) and (2) we get, a = -15 cm · s-2 and u = 115 cm · s-1
Velocity after 7 s of motion, v = u + at = 115 – 15 x 7 = 115 – 105 = 10cm · s-1
Example 3. A man is 9 m behind a train at rest. The train starts with an acceleration of 2 m · s-2 and simultaneously the man starts running. He is able to board the train somehow after 3 seconds. Find the acceleration of the man.
Solution:
While boarding the train, the positions of the man and of the train must be the same.
Let the acceleration of the man be a and the distance traversed by the train in 3 s be x.
We know, s = ut + \(\frac{1}{2}\)at².
Here u = 0 as the train as well as the man starts from rest.
So, for the train, x = \(\frac{1}{2}\) x 2 x (3)² or, x = 9 m
Thus, the distance traversed by the man in this time = 9 + 9 = 18 m
Then, s = 18 m , u = 0 and t = 3 s
∴ a = \(\frac{2 s}{t^2}\) = \(\frac{2}{3 ^2}\) x 18 = 4 m s
Example 4. A particle moves with a uniform acceleration along a straight line. It covers 41 cm and 49 cm in the 6th and the 10th seconds respectively. What will be the distance covered by the particle in 15 seconds?
Solution:
We know, distance covered in the nth second, s = u + \(\frac{1}{2}\)a(2n- 1)
According to the question, s0 = u + \(\frac{1}{2}\)a(2 x 6-1) or, 41 = u + \(\frac{11}{2}\) a….(1)
ans s10 = u + \(\frac{1}{2}\)a(2 x 10- 1) or, 49 = u + \(\frac{19}{2}\)a…..(2)
By solving equations (1) and (2) we get, u = 30 and a = 2
Now, putting t = 15 in s = ut + \(\frac{1}{2}\) at² we get,
s = 30 x 15 = \(\frac{1}{2}\) x 2 x (15)² = 450 + 225 = 675 cm
∴ The particle traverses 675 cm in 15 s.
Example 5. A train begins its journey from station A and stops at station B after 45 min. C is a certain point between A and B where the train attains its maximum velocity of 50 km · h-1. If the train travels from A to C with a uniform acceleration and from C to B with a uniform retardation, calculate the distance between A and B.
Solution:
Let the train start from A with a uniform acceleration a1 and reach C in time t1. From here it travels to B in time t2 with a uniform retardation a2.
The maximum velocity of A the train is v at point C.
Let AC = s1 and CB = s2
So for the motion from A to C, v = \(a_1 t_1 \text { and } s_1=\frac{1}{2} a_1 t_1^2\)
and for the motion from C to B, \(0=v-a_2 t_2 \quad \text { or, } v=a_2 t_2\)
and \(s_2=v t_2-\frac{1}{2} a_2 t_2^2=a_2 t_2^2-\frac{1}{2} a_2 t_2^2=\frac{1}{2} a_2 t_2^2\)
So the distance between A and B,
s = \(A B=A C+C B=s_1+s_2\)
= \(\frac{1}{2} a_1 t_1^2+\frac{1}{2} a_2 t_2^2\)
= \(\frac{1}{2} v t_1+\frac{1}{2} v t_2\) (because \(v=a_1 t_1=a_2 t_2\))
= \(\frac{1}{2} v\left(t_1+t_2\right)=\frac{1}{2} \times 50 \times \frac{3}{4}\)
(\(v=50 \mathrm{~km} \cdot \mathrm{h}^{-1}, t_1+t_2=45 \mathrm{~min}=\frac{3}{4} \mathrm{~h}\))
=18.75 km
Example 6. A train moving with constant acceleration crosses an observer standing on the platform. The first and the second compartments, each 15 m long, cross the observer in 2 s and 2.5 s, respectively. Find the velocity of the train when its first compartment just crosses the observer and also find its acceleration.
Solution:
Let the velocity and the acceleration of the train as its 1st compartment just reaches the observer be u and a, respectively.
Hence, displacement in 2 s = length of the 1st compartment = 15 m
and displacement in (2 + 2.5) or, 4.5 s = total length of the two compartments = 2 x 15 = 30 m
Now from the equation s = ut + \(\frac{1}{2}\) at² we get,
∴ 15 = 2u + \(\frac{1}{2}\)a(2)²
or, u + a = \(\frac{15}{2}\)….(1)
and 30 = 4.5 u + \(\frac{1}{2}\) a(4.5)²
or, 36u + 81a = 240….(2)
Solving equations (1) and (2), we get, u = \(\frac{49}{6}\) and a = –\(\frac{2}{3}\)
The 1st compartment crosses the observer in 2 s; the velocity at that moment,
v = u + at = \(\frac{49}{6}\) x 2 = \(\frac{41}{6}\)
Therefore, the velocity and acceleration of the train as its 1st compartment just crosses the observer are \(\frac{41}{6}\) m · s-1 and –\(\frac{2}{3}\)m · s-2 respectively.
Example 7. A bullet with an initial velocity u penetrates a target. After penetrating a distance x, its velocity decreases by \(\frac{u}{n}\). How much farther will the bullet move through the target before it comes to rest?
Solution:
Let us assume that the retardation of the bullet inside the target is a and it is uniform.
After penetrating a distance x, the velocity, v = u – \(\frac{u}{n}\) = \(\frac{u(n-1)}{n}\)
∴ From \(v^2=u^2-2 a s\), we get, \(\frac{u^2(n-1)^2}{n^2}=u^2-2 a x\)
or, \(2 a x=u^2-\frac{u^2(n-1)^2}{n^2}\)
= \(u^2\left(1-\frac{n^2-2 n+1}{n^2}\right)=u^2 \cdot \frac{2 n-1}{n^2}\)
or, a = \(\frac{u^2(2 n-1)}{2 x n^2}\)
Let the bullet travel an additional distance y before it comes to rest.
∴ \((0)^2=u^2-2 a(x+y) \quad \text { or, } x+y=\frac{u^2}{2 a}\)
or, \(y=\frac{u^2}{2 a}-x=\frac{u^2 \cdot 2 x n^2}{2 u^2(2 n-1)}-x=\frac{x n^2}{2 n-1}-x\)
= \(x\left(\frac{n^2-2 n+1}{2 n-1}\right)=\frac{x(n-1)^2}{2 n-1}\)
Example 8. Starting from rest, a train travels a certain distance with a uniform acceleration α. Then it travels with a uniform retardation β and finally comes to rest again. If the total time of motion is t, find
- The maximum velocity attained and
- The total distance travelled by the train.
Solution:
1. Let t1 be the time taken to travel a distance s1 with acceleration α, and t2 be the time taken to travel a farther distance s2 with retardation β. Let the maximum velocity attained by the train be v.
Here t = t1 + t2
∴ For the motion of the train with acceleration α,
v = \(\alpha t_1 \quad \text { or, } t_1=\frac{\nu}{\alpha}\)…..(1)
and \(\nu^2=2 \alpha s_1 or, s_1=\frac{\nu^2}{2 \alpha}\)…..(2)
Similarly, for the motion of the train with retardation β,
0 = \(\nu-\beta t_2\)
or, \(t_2=\frac{v}{\beta}\) and \(0^2=v^2-2 \beta s_2\) or, \(s_2=\frac{\nu^2}{2 \beta}\)
From (1) and (3) we get, \(t=t_1+t_2=v\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \quad \text { or, } \quad v=\frac{\alpha \beta t}{\alpha+\beta}\)
2. From (2) and (4), the total distance travelled,
s = \(s_1+s_2=\frac{\nu^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \)
= \(\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^2 \times \frac{1}{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right)=\frac{\alpha \beta t^2}{2(\alpha+\beta)} .\)
Example 9. A particle travelling with uniform acceleration along a straight line has average velocities v1, v2 and v3 in successive time intervals t1, t2 and t3, respectively. Prove that, \(\frac{v_2-v_1}{v_3-v_2}=\frac{t_1+t_2}{t_2+t_3}\)
Solution:
Let the initial velocity of the particle be w and its acceleration be a. Also, x, y and z are the velocities after the respective time intervals t1, t2 and t3.
∴ \(x=u+a t_1, y=u+a\left(t_1+t_2\right), z=u+a\left(t_1+t_2+t_3\right) \)
Again, \(v_1=\frac{u+x}{2}, v_2=\frac{x+y}{2}, v_3=\frac{y+z}{2}\)
∴ \(\frac{v_2-v_1}{v_3-v_2}=\frac{\frac{1}{2}(x+y)-\frac{1}{2}(u+x)}{\frac{1}{2}(y+z)-\frac{1}{2}(x+y)}\)
= \(\frac{y-u}{z-x}=\frac{u+a\left(t_1+t_2\right)-u}{u+a\left(t_1+t_2+t_3\right)-\left(u+a t_1\right)}\)
= \(\frac{t_1+t_2}{t_2+t_3}\)
Example 10. A bullet, moving with a velocity of 200 m • s-1 can just go through a 4 cm thick plank. What should be the velocity of a bullet for just going through a 10 cm thick identical plank?
Solution:
The retardation (a) should be the same inside both the planks. The final velocity of the bullet in both cases is zero.
Let the initial velocities of the bullet in the two cases be u1 and u2, respectively.
Then 0 = \(u_1^2-2 a s_1 or, u_1^2=2 a s_1\)
Similarly, \(u_2^2=2 a s_2\)
∴ \(\frac{u_1^2}{u_2^2}=\frac{s_1}{s_2}\)
or, \(u_2=u_1 \sqrt{\frac{s_2}{s_1}}=200 \times \sqrt{\frac{10}{4}}\)
= \(100 \sqrt{10}=316.2 \mathrm{~m} \cdot \mathrm{s}^{-1}\).
Example 11. The speed of a train drops from 48 km h-1 to 24 km h-1 after moving through a distance of 108 m with uniform retardation. How much farther would it move with the same retardation before coming to rest?
Solution:
⇒ \(v^2=u^2+2\) as; \(s=108 \mathrm{~m}=0.108 \mathrm{~km}\)
∴ \(a=\frac{v^2-u^2}{2 s}\)
= \(\frac{(24)^2-(48)^2}{2 \times 0.108}=-8000 \mathrm{~km} \cdot \mathrm{h}^{-2}\)
For the second part of the motion, u = \(24 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { and } v=0\)
∴ \(s=\frac{v^2-u^2}{2 a}\)
= \(\frac{0-(24)^2}{2 \times(-8000)}=0.036 \mathrm{~km}=36 \mathrm{~m}\)
Example 12. A particle starts with a velocity u with a uniform acceleration f. In the p-th, q-th and r-th seconds, it moves through distances a, b and c respectively. Prove that, a(q- r) + b(r- p) + c(p- q) = 0.
Solution:
Distance travelled in the n-th second, \(s_n=u+\frac{1}{2} f(2 n-1)\)
∴ a = \(u+\frac{1}{2} f(2 p-1)\)
= \(u+f p-\frac{f}{2}=\left(u-\frac{f}{2}\right)+f p\)…..(1)
Similarly, b = \(\left(u-\frac{f}{2}\right)+f q\)…..(2)
and c = \(\left(u-\frac{f}{2}\right)+f r\)……(3)
Multiplying (1), (2) and (3) by (q-r),(r-p) and (p-q) respectively, and then adding, we get a(q-r)+b(r-p)+c(p-q)
= \(\left(u-\frac{f}{2}\right)[(q-r)+(r-p)+(p-q)]\) + \(f[p(q-r)+q(r-p)+r(p-q)]\)
= \(\left(u-\frac{f}{2}\right) \cdot 0+f \cdot 0=0\)
Example 13. From two stations A and B, two trains started simultaneously towards each other with velocities v1 and v2 respectively. After they crossed each other, the first train reached B in time t1 and the second train reached A in time t2. Show that v1: v2 = √t2:√t1
Solution:
Let the two trains cross each other at point O after time t.
So, AO = v1t and BO = v2t, i.e., \(\frac{A O}{B O}=\frac{v_1}{v_2}\)
Again, for the first train, OB = v1t1
and for the second train, OA = v2t2
∴ \(\frac{O A}{O B}=\frac{A O}{B O}=\frac{v_2 t_2}{v_1 t_1}\)
Then, \(\frac{v_1}{v_2}=\frac{v_2 t_2}{v_1 t_1} \quad or, \frac{v_1^2}{v_2^2}=\frac{t_2}{t_1} \quad or, \frac{v_1}{v_2}=\frac{\sqrt{t_2}}{\sqrt{t_1}}\)
Example 14. A train attains a velocity v after starting from rest with a uniform acceleration α. Then the train travels for some time with uniform velocity, and at last, comes to rest with a uniform retardation β. If the overall displacement is s in time t, show that \(t=\frac{s}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\).
Solution:
For the accelerated motion, \(\nu=0+\alpha t_1 \quad \text { or, } t_1=\frac{\nu}{\alpha}\)
and \(v^2=0+2 \alpha s_1 \quad or, s_1=\frac{v^2}{2 \alpha}\)
For the uniform motion, \(s_2=v t_2 or, t_2=\frac{s_2}{v}\)
For the retarded motion, \(0=v-\beta t_3 \quad or, t_3=\frac{\nu}{\beta}\)
and 0 = \(v^2-2 \beta s_3 \quad or, s_3=\frac{v^2}{2 \beta}\)
∴ \(s=s_1+s_2+s_3=\frac{\nu^2}{2 \alpha}+s_2+\frac{\nu^2}{2 \beta}\)
or, \(s_2=s-\frac{v^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
Then, t = \(t_1+t_2+t_3=\frac{\nu}{\alpha}+\frac{s_2}{\nu}+\frac{\nu}{\beta}=\frac{v}{\alpha}+\frac{1}{\nu}\left[s-\frac{v^2}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\right]+\frac{\nu}{\beta}\)
= \(v\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\frac{s}{v}-\frac{\nu}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{s}{v}+\frac{\nu}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
Graphical Proof Of The Equations Of Motion: AB represents the velocity-time graph of a particle starting with an initial velocity u, attaining a final velocity v in time t, moving with a uniform acceleration a.
Along the time and the velocity axes, OC = t, OA = u and CB = v.
∴ a = slope of AB
= tan θ= \(\tan \theta=\frac{D B}{A D}=\frac{C B-C D}{A D}=\frac{C B-O A}{O C}=\frac{\nu-u}{t} .\)
Proof Of v = u + at: We know the equation of a straight line of slope m with an intercept c on the y-axis is y = mx + c …..(1)
Comparing the corresponding values for AB, we get, y = v, m = a, x = t and c = u.
∴ The graph AB follows the equation v = u + at….(2)
Proof Of s = ut + \(\frac{1}{2}\)at²: The area under the graph AB and the time-axis, gives the displacement of the particle in time t.
The area under AB is the area of the trapezium OABC, which is the sum of the areas of the rectangle OADC and the triangle ABD.
Hence, the displacement of the particle in time t,
s = \(O A \times O C+\frac{1}{2} A D \times D B=O A \times O C+\frac{1}{2} \times \frac{D B}{A D} \times A D^2\)
∴ \(s=u t+\frac{1}{2} a t^2\) (because A D=O C=t) ……(3)
Proof Of v² = u² + 2as: Slope of AB = a = tanθ = \(\frac{D B}{A D}\)=\(\frac{DB}{O C}\)
∴ as = slope of AB X area of the trapezium OABC
= \(\frac{B D}{O C} \times \frac{1}{2}(O A+C B) \times O C\)
= \(\frac{1}{2} B D(O A+C B)=\frac{1}{2}(C B-C D)(O A+C B)\)
∴ as = \(\frac{1}{2}(v-u)(v+u)=\frac{1}{2}\left(v^2-u^2\right)\)
∴ 2as =\(v^2-u^2\)
or, \(v^2=u^2+2 a s\)
Acceleration in One Dimensional Motion Explained
Proof Of st = u + \(\frac{1}{2}\)a(2t-1): Let AP represent the velocity-time graph of a particle starting with an initial velocity u and moving with a uniform acceleration a.
Let points D and E represent the times (t – 1) and t respectively in the motion. Hence, the area of the trapezium BCED denotes the displacement st in the t th second.
∴ st = \(\frac{1}{2}\) (DB + EC) x DE
DE = OE- OD = t- (t- 1) = 1 s
DB = the velocity attained in time (t- 1) = u+ a(t- 1)
EC = velocity attained in time t = u+ at.
∴ st = \(\frac{1}{2}\)[{u+ a(t—1)} + (u+ at)] x 1
= u + \(\frac{1}{2}\) a(2t- 1)….(5)
Derivation Of The Equations Of Motion Using Calculus
Derivation of v = u + at: From definition, acceleration is the rate of change of velocity,
i.e., a = \(\frac{dv}{dt}\) or, dv = a dt
For motion with a uniform acceleration, a = constant.
∴ ∫dv = a∫dt
or, v = at + A [A = integration constant]……(1)
At t = 0, v = u. Then from equation (1), we get,
u = a·0 + A or, A = u
Hence, equation (1) becomes, v = u + at
Derivation of s = ut + \(\frac{1}{2}\) at²: We know the velocity of a particle is the rate of change of its displacement with time.
∴ v = \(\frac{ds}{dt}\) or, \(\frac{ds}{dt}\) = u + at
or, ds = u dt+ at dt, a and u are constants
∫ds = u∫dt+ a∫t dt
or, s = ut + \(\frac{1}{2}\) at² + A [A = integration constant]……(2)
From initial condition, at t = 0 , s = 0
∴ A = 0
∴ Equation (2) becomes, s = ut + \(\frac{1}{2}\) at²
Derivation Of v² = u² + 2as: Velocity of the particle,
v = \(\frac{ds}{dt}\) and acceleration, a = \(\frac{dv}{dt}\) \(\frac{ds}{dt}\) = v\(\frac{dv}{ds}\)
or, a ds = v dv, where a is a constant, a∫ds = ∫v dv
or, as = \(\frac{v^2}{2}\) + B [B = integration constant]…..(3)
At the start of motion, v = u and s = 0
∴ 0 = \(\frac{u^2}{dt}\) + B or, B = –\(\frac{u^2}{2}\)
Substituting this in equation (3), we get,
as = \(\frac{v^2}{2}\) – \(\frac{u^2}{2}\)
or, v² = u² + 2as…..(4)
Derivation Of The Equations Of Motion Using Calculus Numerical Examples
Example 1. s-t graph for a particle, moving with a constant acceleration, subtends 45° angle with the time axis at time t. That angle becomes 60° 1 s later. Find the acceleration of the particle.
Solution:
Let the velocity of the particle at P1 and P2 be u and v respectively.
∴ u = tanθ1 and v = tanθ2
∴ Acceleration of the particle
a = \(\frac{\nu-u}{t_2-t_1}\)
= \(\frac{\tan \theta_2-\tan \theta_1}{t_2-t_1}\)
= \(\frac{\tan 60^{\circ}-\tan 45^{\circ}}{1}\)
= \(\sqrt{3}-1=0.732 \text { unit. } \mathrm{s}^{-2} .\)
Equations of Motion for One Dimensional Movement
Example 2. Displacement x and time t, in a rectilinear motion of a particle, are related as t= √x+3. Here x is measured in metres and t in seconds. Find the displacement of the particle when its velocity is zero.
Solution:
Given, t = √x + 3 or, √x = t-3 or, x = (t-3)².
Velocity, v = \(\frac{dx}{dt}\) = 2(t – 3)
If velocity =0, 2(t – 3) = 0 or, t = 3 s
∴ At t = 3 s , the displacement, x = (3 -3)² = 0
Hence at zero velocity, the displacement is also zero
Example 3. A body starts from rest and moves with an acceleration proportional to time,
- Find its velocity n s after starting,
- What distance will it travel in ns?
Solution:
According to the problem, a ∝ t
or, a = kt, where k is the constant of proportionality.
Now, a = \(\frac{dv}{dt}\)
dv = kt dt or, ∫dv = k∫t dt
∴ v = \(\frac{k t^2}{2}\) = A(where A = integration constant) ……(1)
Also, as v = 0 at t = 0 , we get A = 0
∴ From equation (1 ) we get, v = \(\frac{k t^2}{2}\)….(2)
Again, if the displacement of the object is s, then v = \(\frac{ds}{dt}\)
∴ From equation (2), \(\)
On integration we get,
s = \(\frac{k t^3}{6}\) + B [where B = integration constant]
Also, at t = 0 , s = 0
∴ B = 0
∴ s = \(\frac{k t^3}{6}\)….(3)
At t = ns, from equations(2) and (3) we get,
- v = \(\frac{k n^2}{2}\)
- s = \(\frac{k n^2}{6}\)
Example 4. The velocity of a moving particle v decreases with its displacement. Given, v = v1 – ax where v0 = initial velocity, x = displacement and a is a constant. How long will the particle take to reach point B on the x-axis at a distance xm from the origin?
Solution:
Given, v = v0 – ax.
At the starting point, v = v0.
∴ v0 = v0– ax or, x= 0 [a = constant]
Hence, the particle was initially at the origin.
Now, \(v=\frac{d x}{d t}=v_0-\alpha x\)
or, \(\frac{d x}{v_0-\alpha x}=d t or, \frac{-\alpha d x}{v_0-\alpha x}=-\alpha d t\)
Then, \(\int \frac{-\alpha d x}{v_0-\alpha x}=-\alpha \int d t or, \log _e\left(\nu_0-\alpha x\right)=-\alpha t+c\); [c is the integration constant]
At t=0 (initially), x=0
∴ \(c=\log _e v_0\)
Therefore, \(\log _e\left(v_0-\alpha x\right)=-\alpha t+\log _e v_0\)
or, \(t=\frac{1}{\alpha} \log _e \frac{v_0}{v_0-\alpha x}\)
At the point \(B, x=x_m\).
Let the corresponding time be \(t_m\).
∴ \(t_m=\frac{1}{\alpha} \log _e \frac{v_0}{v_0-\alpha x_m} \text {. }\)
Example 5. The relation between the time taken and the displacement of a moving body is s = 2t²- 3t² + 4t³, where the unit of s is in metres and that of t is in seconds. Find out the displacement, velocity and acceleration of the body 2 s after initiation of the journey.
Solution:
Here, s = 2t – 3t³ + 4t³……(1)
∴ In 2 s the displacement of the body is,
s = (2 x 2)- [3 x (2)²] + [4 x (2)³]
= (4-12 + 32) = 24 m
Now, the velocity is, v = \(\frac{ds}{dt}\) = 2 – 6t + 12t²
∴ After 2s, v = 2- (6 x 2) +[12 x (2)²] = 38 m · s-1
Again, the acceleration is, a = \(\frac{dv}{dt}\) -6 + 24t
∴ Acceleration after 2 s, a = – 6 +.24 X 2 = -6 + 48 = 42 m · s-2.
Example 6. For a particle travelling along a straight line, the equation of motion is s = 16t + 5t². Show that it will always travel with uniform acceleration.
Solution:
Here, s = 16t + 5t²
∴ Velocity, v = \(\frac{ds}{dt}\) = 16 + 10t
Again acceleration, a = \(\frac{dv}{dt}\) = 10 = constant
∴ The particle will always travel with uniform acceleration.
Example 7. If a, b and c are constants of motion and s = at² + bt + c, then prove that 4a(s – c) = v² – b².
Solution:
Here, s = at² + bt+ c
∴ v = \(\frac{ds}{dt}\) = 2at+ b
∴ v² – b² = (2at+ b)²- b² = 4a²t² + 4abt
= 4a(at² + bt) = 4a(at² + bt + c – c)
or, v² – b² = 4a(s-c) (Proved)
Example 8. The retardation of a particle in rectilinear motion is proportional to the square root of its velocity v. Assume that the constant A of proportionality Is positive. The initial velocity of the particle is v0. How far would the particle move before coming to rest? What would be the time required to travel that distance?
Solution:
⇒ \(a \propto-\sqrt{v}\)
∴ \(a=-A \sqrt{v}=-A \nu^{1 / 2}\)
or, \(\frac{d \nu}{d t}=-A v^{1 / 2}\)
or, \(v^{-1 / 2} d v=-A d t\)
∴ \(\int v^{-1 / 2} d v=-A \int d t+c\) [c=integration constant]
or, \(2 v^{1 / 2}=-A t+c\)…..(1)
Given, at t=0, v=\(v_0\). Putting in (1), \(c=2 v_0^{1 / 2}\)
∴ \(2 v^{1 / 2}=-A t+2 v_0^{1 / 2}\)
or, \(2\left(\sqrt{v_0}-\sqrt{v}\right)\)= At…..(2)
When the particle comes to rest after a time T, we have v=0 at t=T.
From (2), \(2 \sqrt{v_0}=A T \quad \text { or, } T=\frac{2}{A} \sqrt{v_0}
\)
Now, \(a=\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=\frac{d v}{d x} v\)
or, \(d x=\frac{1}{a} v d v=-\frac{1}{A \sqrt{v}} v d v=-\frac{1}{A} v^{1 / 2} d v\)
∴ \(\int d x=-\frac{1}{A} \int v^{1 / 2} d v+k\) [k = integration constant]
or, \(x=-\frac{1}{A} \cdot \frac{2}{3} \nu^{3 / 2}+k=-\frac{2}{3 A} v^{3 / 2}+k\)….(3)
At start, x=0 and \(v=v_0\), Putting in (3),
0 = \(-\frac{2}{3 A} v_0^{3 / 2}+k \quad \text { or, } k=\frac{2}{3 A} v_0^{3 / 2}\)
So, equation (3) becomes, \(x=\frac{2}{3 A}\left(v_0^{\frac{3}{2}}-v^{\frac{3}{2}}\right)\)……..(4)
When the particle comes to rest, v=0. Then the total distance travelled is, \(x_0=\frac{2}{3 A} v_0^{3 / 2}\)
Example 9. The acceleration-time graph of a particle starting from rest is given. Draw the corresponding velocity-time graph and hence find out the displacement in 6s.
Graphical Representation of One Dimensional Motion
Solution:
In the intervals (0 →1), (2 → 3) and (4 → 5) seconds, acceleration is zero, i.e., velocity = constant.
Again, in the intervals (1 → 2), (3 → 4) and (5 → 6) seconds, the velocity increases uniformly and rises to 1 m/s, 2 m/s and 3 m/s respectively, because the uniform acceleration in each interval is 1 m/s²
∴ The velocity-time graph of the motion is ABCDEF
Displacement in 6 s = area under ABCDEF
= area of 6 unit squares + area of 3 triangles
= 6 x (1 x 1) + 3 x (\(\frac{1}{2}\) x 1 x 1) = 7.5 m
Vertical Motion Under Gravity
Acceleration Due To Gravity: when an object is released from a certain height above the earth’s surface, it moves with vertically downward acceleration. Again, when an object is thrown vertically upwards from the ground, it moves up with a deceleration. An upward deceleration is equivalent to a downward acceleration.
Actually, the acceleration is always downward, and its magnitude is the same for both downward and upward motions.
So, the equations of vertical motion are,
v = u + at ….(1)
h = ut + \(\frac{1}{2}\)at² ……(2)
and v² = u² + 2ah ……(3)
Here, initial velocity is u, velocity after time r is v, vertical displacement in time t is h, and acceleration is a.
This acceleration a is called the acceleration due to gravity or free fall acceleration and is represented by the letter g.
The direction of g is always vertically downwards.
If the downward direction is taken as positive, i.e., a = g, then we get the following equations of motion:
v = u + gt …..(4)
h = ut + \(\frac{1}{2}\)gt² …..(5)
and v² = u² + 2 gh …(6))
If the upward direction is taken as positive, i.e., a = we get the following equations: -g, then
v = u – gt …….(7)
h = ut – \(\frac{1}{2}\)gt² …..(8)
and v² = u² – 2gh …..(9)
Maximum Height Attained: When a body is thrown vertically upwards with a velocity u, it momentarily comes to rest on attaining the maximum height and then again starts falling vertically downwards.
Hence, at maximum height H, the velocity of the body, v = 0.
∴ 0 = u² – 2gH [from equation (9)]
or, H = \(\frac{u^2}{2g}\)…….(10)
Time To Reach The Maximum Height: Let the time required to reach the maximum height be T.
∴ 0 = u² – gT [from equation (7)]
or, T= \(\frac{u}{g}\)……(11)
Time Taken To Fall From The Maximum Height: if T1 is the time taken by the body to fall from the maximum height to the initial position then, using equation (5),
H = \(\frac{1}{2} g T_1^2\)[as at maximum height, u = 0]
or, \(T_1^2=\frac{2 H}{g}=\frac{2}{g} \cdot \frac{u^2}{2 g}=\frac{u^2}{g^2}\)
∴ \(T_1=\frac{u}{g}\)….(12)
Hence, the time to reach the maximum height is equal to the time to return to the starting point.
Time Of Flight: it is the total time required for upward and downward motions
T’ = T + T1 = \(\frac{u}{g}\) + \(\frac{u}{g}\) = \(\frac{2u}{g}\)
Time Of Flight Alternative Method: After completion of the upward and downward motions, the displacement becomes zero. Thus, using equation (8),
0 = uT’ – \(\frac{1}{2}\)gT’² or, u = \(\frac{1}{2}\)gT’ or, T’ = \(\frac{2u}{g}\).
Time Taken To Reach A Certain Height: Let the object reach a height h at time f. The initial upward velocity = u.
∴ h = \(u t-\frac{1}{2} g t^2\)
or, \(\frac{1}{2} g t^2-u t+h=0\)
or, \(t=\frac{u}{g} \pm \frac{\sqrt{u^2-2 g h}}{g}\)
From this equation, two different values of t are obtained. This is because the object crosses the point at a height h twice, first during the upward motion and then during the downward motion.
Velocity At Any Height: From the equation \(v^2=u^2-2 g h, \text { we get, } v= \pm \sqrt{u^2-2 g h} \text {. }\).
The positive (+) sign is applicable for upward motion and the negative (-) sign is applicable for downward motion. So, an object crosses any point with the same magnitude of velocity in its upward and downward motions.
Velocity Of Projection And Velocity Of Return: Let u = velocity of projection, v = velocity of return and total time of flight = T’.
∴ From equation (7), we can write,
v = u – gT’ = u – g · \(\frac{2u}{g}\) = -u (T’ = \(\frac{2u}{g}\))
Hence, the upward velocity of projection is equal in magnitude to the downward velocity with which an object hits the ground.
Vertical Motion Under Gravity Numerical Examples
Example 1. A stone is dropped from a height of 19.6 m. What is the time taken by the stone to travel the last metre of the path?
Solution:
In this case u = 0
∴ \(h=\frac{1}{2} g t^2 \quad \text { or, } t=\sqrt{\frac{2 h}{g}}\)
Let t1 and t2 be the time taken by the stone to travel (19.6-1) = 18.6 m and 19.6 m, respectively.
∴ \(t_1=\sqrt{\frac{2 \times 18.6}{9.8}} \text { and } t_2=\sqrt{\frac{2 \times 19.6}{9.8}}\)
or, \(t_1=1.948 \mathrm{~s} \) and \(t_2=2 \mathrm{~s}\)
Hence, the time taken to travel the last metre, \(t_2-t_1=2-1.948=0.052 \mathrm{~s}\)
Example 2. An object is thrown vertically upwards with an initial velocity of 40 m · s-1.
- How long will the object move upwards?
- What will be the maximum height attained?
- How much time will it take to reach the ground?
- When will the object be at a height of 25 m from the ground?
- What will be its velocity after 2 s ? [g = 9.8 m · s-2]
Solution:
1. Let the time taken for upward motion be t s. At maximum height, its velocity is zero. From equation v = u – gt, we get, 0 = 40 – 9.8 x t or, t = 4.1s.
2. Let the maximum height attained be h.
From equation v² = u² – 2gh, we get,
0 = (40)² – 2 x 9.8h or, h = 81.6 m .
3. Let the time taken to reach the ground be t1 starting from the time of projection. Considering both upward and downward motions of the body and using the equation h = ut – \(\frac{1}{2}\)gt², we get,
0 = \(40 t_1-\frac{1}{2} \times 9.8 \times t_1^2\) [total displacement is zero in this case]
∴ t1 = 8.2 s
4. Let the time after which the body is at a height of 25 m be x.
Hence, from h = ut – \(\frac{1}{2}\)gt² we get,
25 = 40 x – \(\frac{1}{2}\) · 9.8x² or, 49x² – 400x + 250 = 0
or, \(x=\frac{400 \pm \sqrt{(400)^2-4 \cdot 49 \cdot 250}}{2 \cdot 49}=\frac{400 \pm 10 \sqrt{1110}}{2 \cdot 49}\)
or, x = 0.682 s and x = 7.481 s
Two values of x signify that the object will be at a height of 25 m twice during its flight, once (x = 0.682 s) while moving upwards and the next (x = 7.481 s) during its downward motion.
5. Let the velocity acquired 2 s after the projection be v.
∴ v = 40 – 9.8 x 2 = 20.4 m · s-1.
Example 3. A ball falls freely on a perfectly elastic plate from a height of 3 m. At the instant t = 0, the velocity of the ball is zero. Draw a velocity-time graph for the motion of the ball, [g = 9.8 m · s-2]
Solution:
In this case, h = \(\frac{1}{2}\) gt²
The time taken by the ball to fall through 3 m is given by t = \(\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 3}{9.8}}=0.783 \mathrm{~s}\)
According to the equation, v² = 2gh, the velocity of the ball just before striking the plate is given by
v = \(\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 3.0}=7.67 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Since the plate is perfectly elastic, the ball after striking it will rebound with the same velocity (7.67 m · s-1) and its velocity will become zero after the same time (0.78 s). This motion will be repeated again and again as shown in the above graph.
Example 4. A body is thrown vertically upwards. After attaining half of its maximum height its velocity becomes 14 m · s-1.
- How high will the body rise?
- What will be the velocity of the body 1 s and 3 s after its projection?
- What is the average velocity of the body in the first half second?
Solution:
1. Let the velocity of projection be u.
Hence, maximum height attained by the body, h = \(\frac{u^2}{2g}\) [using v² = u² – 2 gh]
or, u² = 2 gh
For half the maximum height, i.e., \(\frac{h}{2}\), we get
(14)² = u² – 2g\(\frac{h}{2}\) = 2gh –\(\frac{h}{2}\)(2gh) = gh
∴ h = 20 m [where g = 9.8 m · s-2]
∴ The body will rise up to a height of 20 m. nil Here, the velocity of projection,
2. Here, the velocity of projection, \(u=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 20}=19.8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
∴ The velocity of the body 1 s after projection is, v1 = u – g ·1 = 19.8 – 9.8 = 10 m · s-1.
Velocity 3 s after projection, v2 = u – g x 3 = 19.8 – 9.8 x 3 = -9.6 m · s-1
(Negative sign indicates downward motion of the body.)
3. Velocity after \(\frac{1}{2}\)s
v’ = u – g x \(\frac{1}{2}\) = 19.8 – 9.8 x \(\frac{1}{2}\) = 149 m · s-1
Hence, average velocity during the given period = \(\frac{u+v^{\prime}}{2}=\frac{(19.8+14.9)}{2}=17.35 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)
Short Answer Questions on One Dimensional Motion
Example 5. A piece of stone was dropped from a stationary balloon. The stone covered 13.9 m during the last 1/7 s of its descent. Find the height of the balloon and the velocity of the stone when it strikes the ground, [g = 9.8 m · s-2]
Solution:
Let h = height of the balloon, t = total time of fall of the stone, h’ = downward displacement in time (t – \(\frac{1}{7}\))s.
Hence, \(h-h^{\prime}=\frac{1}{2} g t^2-\frac{1}{2} g\left(t-\frac{1}{7}\right)^2\)
or, \(13.9=\frac{1}{2} \times 9.8 \times t^2-\frac{1}{2} \times 9.8\left(t-\frac{1}{7}\right)^2\)
or, \(13.9=4.9 t^2-4.9 t^2+1.4 t-0.1\)
or, \(1.4 t=14\) or, t=10
∴ Height of the balloon, \(h=\frac{1}{2} g t^2=\frac{1}{2} \times 9.8 \times(10)^2=490 \mathrm{~m}\)
The velocity of the stone when it strikes the ground is, \(v=u+g t=0+9.8 \times 10=98 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)
Example 6. A stone is dropped from the top of a tower 400 m high. At the same time, another stone is thrown upwards from the ground with a velocity of 100 m · s-1. When and where will they meet each other? (g = 9.78 m · s-2).
Solution:
Let the two stones meet after a time t at a distance h from the top of the tower.
h = \(\frac{1}{2}\) gt² = \(\frac{1}{2}\) x 9.8 t² ……(1)
Considering the downward motion of the 1st stone 400 – h = 100t – \(\frac{1}{2}\) x 9.8 t² …..(2)
From equations (1) and (2) we get, 400 – \(\frac{1}{2}\)x 9.8t² = 100t – \(\frac{1}{2}\) x 9.8t²
or, 100t = 400 or, t = 4 s
Hence, from equation (1) we get, h = \(\frac{1}{2}\) x 9.8 x (4)² = \(\frac{1}{2}\) x 9.8 x 16 = 78.4 m
Hence, the two stones meet at 78.4 m below the top of the tower after 4 s.
Example 7. A stone is dropped from the top of a vertical pillar. When the stone has fallen through a height x, another stone is dropped from height y below the top of the pillar. Both the stones touch the ground at the same time. Prove that the height of the pillar should be \(\frac{(x+y)^2}{4 x}\).
Solution:
Let the height of the pillar be h and the velocity of the stone at x below the top of the pillar be v.
∴ \(v^2=2 g x \text { or, } v=\sqrt{2 g x}\)…..(1)
Let the first stone take \(t \mathrm{~s}\) to cover the distance (h-x).
∴ \(h-x=v t+\frac{1}{2} g t^2\)….(2)
According to the problem, the second stone is dropped from a height of (h-y) and this stone takes time t to cross that distance.
∴ \(h-y=\frac{1}{2} g t^2\)…..(3)
From equations (2) and (3) we get, \(h-x=\nu t+h-y \text { or, } y-x=v t \text { or, } y-x=t \sqrt{2 g x}\)
∴ t = \(\frac{y-x}{\sqrt{2 g x}}\)
From equation (3) we get, \(h-y=\frac{1}{2} g \frac{(y-x)^2}{2 g x}=\frac{(y-x)^2}{4 x}\)
or, \((h-y) \cdot 4 x=(y-x)^2 or, 4 x h-4 x y=y^2-2 y x+x^2\)
or, \(h=\frac{(x+y)^2}{4 x}\) (Proved).
Example 8. A, B, C and D are four points on a vertical line such that AB = BC = CD. A body is allowed to fall freely from A. Prove that the respective times required by the body to cross the distances AB, BC, and CD should be in the ratio 1: (√2- 1): (√3-√2).
Solution:
Let AB – BC = CD = x and the time taken by the body to cover these distances be t1, t2 and t3 respectively.
Now, \(x=\frac{1}{2} g t_1^2 \text { or, } t_1=\sqrt{\frac{2 x}{g}}\)…..(1)
2x = \(\frac{1}{2} g\left(t_1+t_2\right)^2 \text { or, } t_1+t_2=\sqrt{\frac{4 x}{g}}\)…..(2)
3 x = \(\frac{1}{2} g\left(t_1+t_2+t_3\right)^2 \text { or, } t_1+t_2+t_3=\sqrt{\frac{6 x}{g}}\)…..(3)
From (1) and (2) we get, \(t_2=\sqrt{\frac{4 x}{g}}-\sqrt{\frac{2 x}{g}}=\sqrt{\frac{2 x}{g}}(\sqrt{2}-1)\)
From (2) and (3) we get, \(t_3=\sqrt{\frac{6 x}{g}}-\sqrt{\frac{4 x}{g}}=\sqrt{\frac{2 x}{g}}(\sqrt{3}-\sqrt{2})\)
∴ \(t_1: t_2: t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})\) (Proved).
Example 9. A rubber ball is thrown vertically downwards from the top of a tower with an Initial velocity of 14 m · s-1. A second ball is dropped 1 s later from the same place. In 2 s the first ball reaches the ground and rebounds upwards with the same velocity. When will they collide with each other?
Solution:
Height of the tower, h = distance covered by the first ball in 2s = 14 x 2 + \(\frac{1}{2}\) x 9.8 x (2)² = 47.6 m
(h = ut + \(\frac{1}{2}\)gt²)
The velocity of the first ball just before touching the ground is v = 14 + 9.8 x 2 = 33.6 m · s-1
Hence, its velocity just after bouncing = 33.6 m · s-1
Downward displacement of the second ball in 1 s, x = \(\frac{1}{2}\) x 9.8 x (1)² = 4.9 m
Velocity of second ball after 1 s = 9.8 x 1 = 9.8 m · s-1
Hence, the distance between the two balls, 2 s after the projection of the first ball = 47.6- 4.9 = 42.7 m.
Let the two balls collide with each other ts after the first ball bounces off the ground.
Upward displacement of the first ball in ts, x1 = 33.6t – \(\frac{1}{2}\) X 9.8 x t² = 33.6t- 4.9t²
Downward displacement of the second ball in t s, x2 = 9.8t+ \(\frac{1}{2}\) x 9.8 x t² = 9.8t + 4.9t²
Now, x1 + x2 = 42.7
or, 33.6t – 4.9t + 9.8t+ 4.9 t³ = 42.7 or, 43.4t = 42.7
∴ t = \(\frac{42.7}{43.78}\) = 0.98 s
Example 10. A lift starts to move up with a constant acceleration of 2 m · s-2 from the earth’s surface. A piece of stone is dropped outside from the lift 4 s after the start of the lift When will the stone reach the earth’s surface?
Solution:
Initial velocity of the lift, u = 0; acceleration, a = 2 m · s-2
Let the rise of the lift in 4 s be s and its velocity at that point be v.
Hence, from equation v = u+ at we get,
v = 0 + 2 x 4 = 8 m · s-1
Also from equation s = ut + \(\frac{1}{2}\) at² we get
s = 0 x 4 + \(\frac{1}{2}\) x 2 (4)² = 16 m
∴ The stone piece was dropped with an initial upward velocity of 8 m · s-1 and was at a height of 16 m from the ground. If t is the time taken by the stone to reach the ground, then from equation h = ut + \(\frac{1}{2}\)gt²,
16 = \(-8 t+\frac{1}{2} \times 9.8 t^2\)
(because for the stone, \(u=-8 \mathrm{~m} \cdot \mathrm{s}^{-1}, g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\), h=16 m)
or, \(4.9 t^2-8 t-16=0\)
or, \( t=\frac{8 \pm \sqrt{(8)^2-4 \times 4.9 \times(-16)}}{2 \times 4.9}=\frac{8 \pm 19.4}{9.8}\)
Since time cannot be negative, \(t=\frac{8+19.4}{9.8}=2.8 \mathrm{~s}\).
Example 11. Two bodies released from different heights fall freely and reach the ground at the same time. The first body takes a time, t1 = 2 s and the second body takes a time, t2 = 1 s. What was the height of the first body at the time of the release of the second body?
Solution:
Let the height from which the first body is released be h1 above the ground. When it is at a height h2, the second body is released.
So, the first body falls through a height (h1 – h2) in time (t1– t2) before the release of the second body.
According to the problem, \(h_1=\frac{1}{2} g t_1^2 \text { and } h_1-h_2=\frac{1}{2} g\left(t_1-t_2\right)^2\)
or, \(\frac{1}{2} g t_1^2-h_2=\frac{1}{2} g\left(t_1-t_2\right)^2\)
or, \(h_2=\frac{1}{2} g\left(2 t_1-t_2\right) t_2\)
= \(\frac{1}{2} \times 9.8(2 \times 2-1) \times 1=14.7 \mathrm{~m} .\)
Example 12. A balloon moves vertically upwards with a uniform velocity v0. A weight is tied to the balloon with a rope. When the balloon attains a height of h0, the rope snaps. How much time will the weight take to reach the ground?
Solution:
Let the time taken by the weight, after the rope snaps, to reach the ground be t. From the question, the weight ascends with the same velocity as that of the balloon, i.e., v0.
So, when the rope snaps the initial velocity of the weight =- v0 [the negative sign comes as the velocity of the weight is directed upwards just when the rope snaps].
∴ For the free fall of the weight, \(h_0=-v_0 t+\frac{1}{2} g t^2 \text { or, } \frac{1}{2} g t^2-v_0 t-h_0=0\)
∴ \(t=\frac{v_0 \pm \sqrt{v_0^2-4 \times \frac{1}{2} g\left(-h_0\right)}}{2 \cdot \frac{1}{2} g}=\frac{v_0 \pm \sqrt{v_0^2+2 g h_0}}{g}\)
So, the time taken by the weight to reach the ground is \(\frac{v_0+\sqrt{v_0^2+2 g h_0}}{g}\) (becasue t>0).
Example 13. According to the, three cars P, Q and R are at three points along the x-axis at a given moment. Now car P starts its motion towards P1 parallel to the y-axis with a uniform velocity v. Again, R is in motion parallel to the y-axis along RRX with uniform acceleration a. If car Q too moves parallel to the y-axis then under what condition will all of them remain collinear? Given PQ = QR
Solution:
Let the time be t after which P and R are at and respectively.
PP1 = vt and RR1 = \(\frac{1}{2}\)at²
According to the question, after time t, Q will be at Q1.
From the figure we have, \(\frac{P P_1}{P M}=\frac{Q Q_1}{Q M}=\frac{R R_1}{R M}=k\) (say)
∴ \(P P_1=k \cdot P M=k(P Q+Q M)\)
⇒ \(Q Q_1=k \cdot Q M\)
and \(R R_1=k \cdot R M=k(Q R-Q M)\)
∴ \(P P_1-R R_1=k \cdot(Q M+Q M)\) (because P Q=Q R)
= \(2 k \cdot Q M\)
∴ \(k \cdot Q M=\frac{P P_1-R R_1}{2}\)
∴ \(Q Q_1=k \cdot Q M\)
= \(\frac{P P_1-R R_1}{2}=\frac{P P_1}{2}-\frac{R R_1}{2}=\frac{\nu t}{2}-\frac{1}{4} a t^2\)
∴ \(Q Q_1=\left(\frac{\nu}{2}\right) t-\frac{1}{2}\left(\frac{a}{2}\right) \cdot t^2\)……..(1)
Comparing equation (1) with s = ut- \(\frac{1}{2}\)at², we can say that to remain in the same straight line joining the other two cars, the initial velocity of the car Q must be \(\frac{v}{2}\) along the positive y-axis and it should move with retardation \(\frac{a}{2}\).
One-Dimensional Motion Conclusion
Dynamical quantities related to an object are determined with reference to some other object in the surroundings. This external object forms a frame of reference. A frame of reference is represented in forms like cartesian, polar, spherical etc.
- The change of position of a moving object in a fixed direction is called displacement. Displacement is a vector quantity.
- The rate of Hie distance covered by a body with respect to time is called speed. Speed is a scalar quantity.
- The rate of the distance travelled from a point by a particle with respect to an infinitesimal interval of time is called the instantaneous speed of the particle at that point.
- The rate of displacement of a body with respect to time is called velocity. Velocity is a vector quantity.
Applications of One Dimensional Motion in Real Life
The rate of the displacement of a particle from a certain point with respect to an infinitesimally small time interval is called the instantaneous velocity of the particle at that point.
- An object moving in a straight line with uniform speed is said to be moving with uniform velocity.
- An object with uniform velocity must have uniform speed but an object with uniform speed may not have uniform velocity.
- For a moving object, zero average velocity does not necessarily mean a zero average speed, but zero speed always implies zero velocity.
- For an object moving with a uniform velocity, its average velocity is the same as its instantaneous velocity.
The rate of change of velocity of an object with respect to time is called acceleration. Acceleration is a vector quantity.
- The rate of change in velocity of a body at a point with respect to an infinitesimally small time interval is the instantaneous acceleration of the body at that position.
- On a displacement-time graph of a particle, the inclination of the tangent at a point on the graph gives the instantaneous velocity of the particle at that point.
- In a velocity-time graph for a particle, the inclination of the tangent at any point gives the instantaneous acceleration of the particle at that point.
The area enclosed between the velocity-time graph of a particle and the time axis gives the magnitude of the displacement of the particle.
One-Dimensional Motion Useful Relations For Solving Problems
Speed, v = \(\frac{l}{t}\), where l = distance covered in time t.
Average speed = \(\frac{\text { total distance }}{\text { total time }}\)
or, \(v=\frac{l_1+l_2+l_3+\cdots+l_n}{t_1+t_2+t_3+\cdots+t_n}\)
Instantaneous speed, \(v_i=\lim _{\Delta t \rightarrow 0} \frac{\Delta l}{\Delta t}=\frac{d l}{d t}\)
Velocity, v = \(\frac{s}{t}\), where s= displacement in time t.
Average velocity = \(\frac{\text { total displacement }}{\text { total time }}\)
or, (v)= \(\frac{s_1+s_2+s_3+\cdots+s_n}{t_1+t_2+t_3+\cdots+t_n}\)
Instantaneous velocity, \(v_i=\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t}\)
For A Particle In Motion
- Initial velocity = u
- Acceleration = a
- Distance covered in the n-th second =sn
- Acceleration due to gravity = g
- The final velocity after time t = v
- Displacement in time t = sn
- Height in time t = h
Acceleration or average acceleration = \(\frac{\text { final velocity }- \text { initial velocity }}{\text { time }}=\frac{\text { change in velocity }}{\text { time }}\)
or, (a) =\(\frac{v-u}{t}\)
Instantaneous acceleration, \(a_i=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}=\frac{d^2 s}{d t^2}\)
For a particle in motion with uniform acceleration,
- v = u + at,
- s = ut +\(\frac{1}{2}\)at²,
- v² = u² + 2as,
The displacement in the t th second st = u + \(\frac{l}{2}\)a(2t-1) In case of retardation,
- v – u -at,
- st = ut – \(\frac{1}{2}\)at²,
- v² = u² – 2as,
- st = u – \(\frac{1}{2}\)a(2t-1)
Equations of vertical motion under gravity (downward direction positive):
- v = u + gt,
- h = ut + \(\frac{1}{2}\)gt²,
- v² = u² + 2gh
(upward direction positive):
- v = u – gt,
- h = ut- \(\frac{l}{2}\)gt²,
- v² = u² – 2gh
One-Dimensional Motion Very Short Answer Type Questions
Question 1. A particle is in uniform motion with respect to a reference frame. Is it possible for the particle to be at rest with respect to another frame?
Answer: Yes
Question 2. Retardation is essentially a _______ acceleration
Answer: Negative
Question 3. For a particle, displacement (x) and time (t) are related by the following equation: x = (3t² + 2t+5)m. If time is expressed in seconds, find the initial velocity of the particle.
Answer: 2 m · s-1
Question 4. The displacement of a moving particle is directly proportional to the square of the time duration. State whether the particle is moving at a constant velocity or at a constant acceleration.
Answer: Constant acceleration
Question 5. When is the average velocity of a particle equal to its instantaneous velocity?
Answer: In motion with uniform velocity
Question 6. Does a particle with a uniform speed in a curved path possess any acceleration?
Answer: Yes
Question 7. How many dimensions are there in the motion of a ship in a turbulent sea?
Answer: Three
Question 8. If the position of a particle at instant t is given by x = t4, find the acceleration of the particle.
Answer: 12t²
Question 9.For a moving body, displacement y (in metre) and time t (in second) are related as y = – \(\frac{2}{3}\) t²+16t+2. When will the body stop?
Answer: After 12s
Question 10. The displacement equation for a particle moving in a straight line is x = αt³ + βt² + γt+ δ. The ratio of the initial acceleration to the initial velocity depends only on _______
Answer: β and γ
Question 11. The motion of an artificial satellite around the earth is a _________ dimensional motion.
Answer: Two
Question 12. An athlete runs with a velocity of 18 km · h-1. How much distance will he cover in 10 min?
Answer: 3 km
Question 13. What is the nature of the time-displacement graph for a particle moving with a constant velocity?
Answer: A straight line
Question 14. What does the slope of a position-time graph represent?
Answer: Velocity of a body
Question 15. The area under v-t graph =?
Answer: Displacement of the particle
Question 16. In the same displacement-time graph, two motions are represented by two straight lines having slopes of 45° and 60° respectively. Which line represents a higher velocity and what is the ratio between the first and the second velocities?
Answer: Second: 1: √3
Question 17. Velocity (in m · s-1 unit)-time (in s unit) graph, of a particle moving in a straight line, is a straight line and it is inclined at an angle 45 with the time axis. What is the acceleration of the particle?
Answer: 1 m s-2
One-Dimensional Motion Assertion Reason Type Questions And Answers
These questions have statements 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.
- Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
- Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
- Statement 1 is true, statement 2 is false.
- Statement 1 is false, and statement 2 is true.
Question 1.
Statement 1: The average velocity of a particle may be equal to its instantaneous velocity.
Statement 2: For a given time interval of a given motion, average velocity is single-valued while average speed can have many values.
Answer: 3. Statement 1 is false, statement 2 is true.
Question 2.
Statement 1: A scooter moves towards the north and then moves towards the south with the same speed. There will be no change in the velocity of the scooter.
Statement 2: Velocity is a vector quantity.
Answer: 4. Statement 1 is false, statement 2 is true.
Question 3.
Statement 1: An object can possess acceleration even when it has a uniform speed.
Statement 2: When the direction of motion of an object keeps changing, its velocity also changes with time.
Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
Question 4.
Statement 1: Acceleration of a moving particle can change its direction without any change in the direction of velocity.
Statement 2: If the direction of the change in the velocity vector changes, the direction of the acceleration vector also changes.
Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
Question 5.
Statement 1: The distance between two particles moving with constant velocities always remains constant.
Statement 2: The relative motion between two particles moving with a constant is velocities.
Answer: 4. Statement 1 is false, statement 2 is true.
Question 6.
Statement 1: A particle with zero velocity may have a non-zero acceleration.
Statement 2: A particle comes to rest at the instant of reversing its direction of motion.
Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
Question 7.
Statement 1: A particle moving with uniform accelera¬tion has its displacement proportional to the square of time.
Statement 2: If the motion of a particle is represented by a straight line on the velocity-time graph, its acceleration is uniform.
Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
Question 8.
Statement 1: A freely falling body travels through distances in this ratio 1: 3: 5: 7:…… in successive equal intervals of time (Galileo’s law of odd integers).
Statement 2: In one-dimensional motion, a particle with zero speed may have a non-zero velocity.
Answer: 3. Statement 1 is true, statement 2 is false.
Question 9.
Statement 1: If the average velocity of a body is equal in two successive time intervals, its velocity is a constant.
Statement 2: When a body travels with constant velocity, its displacement is proportional to time.
Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
Question 10.
Statement 1: If two bodies of different masses one dropped simultaneously from the same height, then they touch the ground simultaneously.
Statement 2: The time of flight of a freely falling body is independent of its mass.
Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
Question 11.
Statement 1: The distance between two bodies does not change if they move in the same direction with the same constant acceleration.
Statement 2: Two bodies moving with the same velocity are at rest relative to each other.
Answer: 4. Statement 1 is false, statement 2 is true.
Question 12.
Statement 1: A body is dropped from a height h and another body is thrown simultaneously from the ground with a velocity u in the vertically upward direction. They meet after a time of \(\frac{h}{u}\).
Statement 2: For a body projected in the vertically upward direction, the ascent in the last second is always 4.9 m, whatever the velocity of projection.
Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
One-Dimensional Motion Match Column A And Column B
Question 1. In column 1 velocity-time graphs of a particle moving along a straight line are given and in column 2 the corresponding acceleration-time graphs are given.
Answer: 1. A, 2. E, 3. D, 4. C
Question 2. In column 1 some statements related to various physical quantities are given while in column 2 information about the motion of a particle moving along a straight line are given.
Answer: 1. C, 2. B, 3. D, 4. A
Question 3. The displacement-time curve of a moving particle is given.
Answer: 1. A, 2. C, 3. B, 4. D
Question 4.
Answer: 1. A, B, 2. A, B, 3. C, 4. A, B
Question 5.
Answer: 1. A, 2. B, E, 3. C, 4. D
Question 6.
Answer: 1. A, B, 2. C 3. A, B, D, 4. C
Question 7.
Answer: 1. C, 2. B, 3. D, 4. A
Question 8. For a moving particle, displacement and velocity at time t are s and v, respectively.
Answer: 1. A, 2. A, 3. D, 4. B
Question 9. Two cars start from the origin and move along the x-axis. Their displacements (in m) are related to time (in s) as, xA = 4t+ t² and xB = 2t² + 2t³. Then
Answer: 1. D, 2. B, 3. C, 4. A
One-Dimensional Motion Comprehension Type Questions And Answers
Read the following passages carefully and answer the questions at the end of them.
Question 1. The vertical rise or fall of a particle under gravity is governed by the equations:
- v = u + gt,
- h = ut + \(\frac{1}{2}\)gt², and
- v² = u2² + 2gh, the symbols having their usual meanings. Then for a particle dropped from the top of a tower and falling freely, choose the correct options:
1. The distance covered by it after n seconds is directly proportional to
- n¹
- n
- 2n-1
- 2n²-1
Answer: 1. n²
2. The distance covered in the nth second is proportional to
- n²
- n
- 2n-1
- 2n²-1
Answer: 3. 2n-1
3. The velocity of the body after n seconds is proportional to
- n²
- n
- 2n-1
- 2n²-1
Answer: 2. n
Question 2. Shows the speed-time graph of two cars A and B which are travelling in the same direction over a period of 40 s. Car A travelling at a constant speed of 40 m · s-1, overtakes car B at t = 0. In order to catch up with car A, car B immediately accelerates uniformly for 20 s.
1. Distance travelled by car B in 20 s is
- 800 m
- 750 m
- 1000 m
- 500 m
Answer: 2. 750m
2. What is the acceleration of car B for the first 20 s?
- 1.25 m · s-2
- 3.75 m · s-2
- 2.5 m · s-2
- 5 m · s-2
Answer: 1. 1.25 m · s-2
3. At what time car B overtake car A?
- 12 s
- 50 s
- 18 s
- 25 s
Answer: 4. 25 s
4. What is the distance travelled by car A before car B overtakes it?
- 480 m
- 1000 m
- 800 m
- 1200 m
Answer: 2. 1000 m
5. What is the maximum separation between the two cars during the 40 s interval?
- 150 m
- 480 m
- 390 m
- None of these
Answer: 1. 150 m
Question 3. A ball is thrown up with velocity u = 50 m · s-1 as shown. The origin and positive and negative directions are also indicated in the figure. Neglect air resistance and take g = 10 m · s-2.
1. How much time does the ball take to reach half the maximum height?
- 1.46 s
- 2.5 s
- 3s
- 1.82s
Answer: 1. 1.46 s
2. After how much time will it again cross the same position?
- 7.07 s
- 7.5 s
- 8.18 s
- Would not cross again
Answer: 1. 7.07 s
3. Determine the distance travelled by the ball in the 3rd second of its motion.
- Zero
- 45 m
- 25 m
- 80 m
Answer: 3. 25
One-Dimensional Motion Integer Answer Type Questions
In this type, the answer to each of the questions is a single-digit integer between 0 and 9.
Question 1. The displacement x of a particle moving in one dimension, under the action of a constant force, is related to time t by the equation t = √x + 3, where x is in meters and t is in seconds. Find the displacement (in metres) of the particle when its velocity is zero.
Answer: 0
Question 2. The motion of a body is defined by \(\frac{dv(t)}{dt}\) = 6-3v(t) where v(t) is the velocity (in m/s) of the body at time t (in seconds). If the body was at rest at t = 0, find its velocity (in m/s) when the acceleration is half the initial value.
Answer: 1
Question 3. A balloon is at a height of 40 m and is ascending with a velocity of 10 m · s-1. A bag of 5 kg weight is dropped from it. When will the bag reach the surface of the earth? Given g = 10 m · s-2.
Answer: 4
Question 4. A bike, initially at rest, travels the first 20 m in 4s along a straight line with constant acceleration. Determine the acceleration of the bike in m · s-2. Consider the bike as a particle.
Answer: 5
Question 5. Starting from rest, a particle moving along a straight line attains a speed of 2 m · s-1 in 1.5 s. What is the particle’s speed after an additional 3 s has elapsed assuming that the particle is moving with constant acceleration?
Answer: 6
