WBCHSE Class 11 Physics One-Dimensional Motion Long Answer Questions

One-Dimensional Motion Long Answer Type Questions

Question 1. Can a particle, moving with a uniform speed have a non-uniform velocity?
Answer:

Speed is a scalar quantity having only magnitude. But velocity is a vector quantity having both magnitude and direction. Hence, if a particle having uniform speed changes its direction, its velocity will become non-uniform.

Example: uniform circular motion.

Question 2. Can a particle, moving with a uniform velocity have i a non-uniform speed?
Answer:

If a particle has uniform velocity, both its magnitude and direction remain constant. Speed is a scalar quantity having magnitude only. For uniform velocity, as the magnitude remains unaltered, speed cannot be non-uniform.

Question 3. Even if the average velocity of a body is zero, its average speed may be non-zero—is it possible?
Answer:

The displacement of a particle becomes zero if the particle returns to its starting point after completion of the journey.

Then the average velocity = \(\frac{\text { total displacement }}{\text { total time }}=0\)

Again, average speed = \(\frac{\text { total distance covered }}{\text { total time }}\); as total distance covered is non-zero here, and the average speed is not equal to zero.

Question 4. State whether a particle having an acceleration may have a velocity of constant magnitude.
Answer:

Velocity of constant magnitude may change its direction with time. This results in a change in velocity, as velocity is a vector quantity. Hence there is a non-zero acceleration because acceleration = \(\frac{\text { change in velocity }}{\text { time }}\).

Question 5. Velocity is zero but acceleration is non-zero —is it possible?
Answer:

A body may have zero velocity but non-zero acceleration. For example, when a body is thrown vertically upwards, at the maximum height the body momentarily comes to rest. At this point, the body attains zero velocity but still has an acceleration equal to the acceleration due to gravity directed downward.

Question 6. Can the directions of velocity and acceleration be
Answer:

The velocity and acceleration of a body can be in different directions. For example, if a body moves towards the east with a retardation, its velocity is directed towards the east. But as retardation is negative acceleration, the acceleration is directed towards the west.

WBBSE Class 11 One-Dimensional Motion Long Answer Questions

Question 7. Can there be any change in the direction of the velocity of a body moving under a constant acceleration?
Answer:

The direction of velocity of a body may change even if the body has a constant acceleration. In projectile motion, the path changes at every point with the change in direction of the velocity. But at every point, its acceleration is the acceleration due to gravity and hence, is a constant.

Question 8. What kind of motion is described by the equation; s = s₀ + ut + \(\frac{1}{2}\)at²?
Answer:

The equation describes the linear motion of a particle with a constant acceleration a, the initial displacement and velocity being s₀ and u, respectively.

Question 9. For a body moving with a uniform acceleration, prove that its average velocity is the arithmetic mean of its initial and final velocities.
Answer:

Let u be the initial velocity of a body and a be its uniform acceleration. After time t, the body acquires a velocity v (say) and s is its displacement. Hence, average velocity of the body = \(\frac{s}{t}\).

Again, the arithmetic mean of the initial and the final velocities of the body

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

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

Question 10. Starting from rest, a body moves in a straight line with constant acceleration. Describe the nature of the graph relating the displacement with time.
Answer:

Displacement of the body that starts from rest and moves with a uniform acceleration, after a time t is given by

s = \(\frac{1}{2}\) at² or, t² = \(\frac{2s}{a}\)

This is the equation of a parabola with a vertex at (0, 0) and its axis along the displacement axis

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Example 11. State the nature of the graphs representing motions of a body with uniform velocity, uniform acceleration and uniform retardation respectively in a displacement-time graph.
Answer:

In a displacement-time graph, the slope of the graph = rate of displacement with time = velocity of the body.

For a body moving with uniform velocity, velocity = constant.

Hence, the slope of the graph is constant. And its displacement-time graph is a straight line.

In case of motion with uniform acceleration, velocity increases and hence the slope of the graph also increases. Line 2 represents this motion. This line is a parabola.

On the other hand, in the case of motion with uniform retardation, the velocity decreases, and therefore, the slope of the graph also decreases. Line 3 represents this motion. This line is also a parabola.

Question 12. How can you represent

1. Motion with uniform velocity,
2. Motion with uniform acceleration and
3. Motion with uniform retardation in a velocity-time graph?

Answer:

1. For motion with uniform velocity, the velocity is constant. Hence the representative line for such a motion is a straight line parallel to the time axis. OA represents the magnitude of the uniform velocity.
2. The slope of the line in the velocity-time (v-t) graph represents the acceleration and for motion with uniform acceleration, the slope is constant and positive.
3. For constant retardation, the slope will be negative as represented by line 3.

Graphical Analysis of One-Dimensional Motion

Question 13. Draw the velocity-time graph of a body moving

1. With uniform acceleration,
2. With increasing acceleration and
3. With decreasing acceleration.

Answer:

In a velocity-time graph, the slope of the graph gives the acceleration.

1. Graph in this case is a straight line as the acceleration of the body is uniform, i.e., constant and so the slope of the straight line is also constant as shown.
2. The slope of the v-t graph will gradually increase with increasing acceleration. The graph will be a curved line represented by line 2.
3. On the other hand, the slope decreases with decreasing acceleration. Thus, the graph will be a curved line represented by line 3.

Question 14. In a velocity-time graph how

1. Uniform retardation,
2. Gradually increasing retardation and
3. Gradually decreasing retardation can be represented?

Answer:

In a velocity-time graph, a negative slope represents the rate of decrease in velocity with time.

1. In the case of uniform retardation, the slope is constant and the graph is a straight line with a negative slope as shown by line 1
2. In the case of gradually increasing retardation, the magnitude of the slope increases gradually. Thus the graph is a curved line, as shown by line 2.
3. In case of gradually decreasing retardation the magnitude of slope will decrease gradually. The graph is a curved line (line 3). In each case, OA represents the initial velocity.

Question 15. A body with an initial velocity u and a uniform acceleration covers a distance s in time t and acquires a velocity v. Compare the velocity of the body at half of the distance covered with the velocity at half of the total time of travel.
Answer:

Let the acceleration of the body be a.

v = \(u+a t \text { or, } a=\frac{v-u}{t}\)

∴ Velocity at half of the total time of travel, \(v_t=u+a \frac{t}{2}=u+\frac{t}{2}\left(\frac{v-u}{t}\right)\)

= \(u+\frac{\nu-u}{2}=\frac{\nu+u}{2}\)

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

∴ Velocity \(v_s\) at half of the distance covered, \(v_s^2=u^2+2 a \cdot \frac{s}{2}=u^2+\frac{s\left(v^2-u^2\right)}{2 s}=\frac{u^2+v^2}{2}\)

Now, \(v_t^2-v_s^2=\left(\frac{\nu+u}{2}\right)^2-\left(\frac{\nu^2+u^2}{2}\right)\)

= \(\frac{v^2+u^2+2 u v}{4}-\frac{v^2+u^2}{2}\)

= \(\frac{\nu^2+u^2+2 u v-2 v^2-2 u^2}{4}\)

= \(-\frac{\left(v^2+u^2-2 u v\right)}{4}=-\left(\frac{u-v}{2}\right)^2\) which is a negative quantity.

Hence v_t <v_s, i.e., the velocity of the body at half-distance is greater than the velocity at half-time.

Question 16. State whether anybody with a two-dimensional motion may have an acceleration in one dimension only.
Answer:

A body in a two-dimensional motion may have an acceleration in one dimension only.

Projectile motion is two-dimensional because at any instant, the object has both horizontal and vertical components of velocity. But the acceleration is one-dimensional—the acceleration due to gravity acting always downwards.

Step-by-Step Solutions to One-Dimensional Motion Problems

Question 17. The displacement of a particle during its motion is equal to half of the product of its instantaneous velocity and time. Show that the particle moves with a constant acceleration.
Answer:

Let the displacement be s, instantaneous velocity be v and the time be t.

From the given condition,

s = \(\frac{1}{2} v t=\frac{1}{2} \frac{d s}{d t} t \text { or, } \frac{d s}{s}=2 \frac{d t}{t}\)

Integrating, we get, \(\int \frac{d s}{s}=2 \int \frac{d t}{t}\)

or, \(\log _e s=2 \log _e t+\log _e c\) [where c is the integration constant]

= \(\log _e c t^2\)

∴ s = \(c t^2\)

v = \(\frac{d s}{d t}=2 c t,\)

and acceleration \(=\frac{d \nu}{d t}=2 c=\) constant .

Question 18. When the speed of a car is doubled, the distance required to stop it becomes 4 times—why?
Answer:

Let the retardation produced by applying the brakes be a. Let its initial velocity be u and the distance covered by the car before coming to rest be s.

Hence, 0 = u² – 2as [from equation v² = u² – 2as]

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

If the retardation is constant, then, s ∝ u².

Hence, if u is doubled, s becomes 2² or 4 times.

Question 19. Does the magnitude of a physical quantity depend on the chosen frame of reference?
Answer:

The magnitudes of the physical quantities related to the intrinsic properties like mass, density, number of molecules, etc. are independent of the frame of reference.

But the magnitudes of the physical quantities denoting extrinsic or dynamic properties like position, displacement, speed, velocity, acceleration, etc. depend on the frame of reference. For example, a train moves with a velocity of 45 km · h^-1 with respect to the ground. But with respect to another train, running with the same velocity, this velocity will be zero.

Note: According to Einstein’s special theory of relativity, intrinsic properties like mass, length, time etc. depend on the frame of reference.

Question 20. A particle in motion covers half of a circular path of radius r in time f. Find the average speed and average velocity of the body.
Answer:

The distance covered in this case = πr

∴ Average speed = \(\frac{\pi r}{t}\)

The displacement in this case = 2 r

Hence, average velocity = \(\frac{2r}{t}\)

Question 21. From the top of a tower, one ball is thrown vertically upwards and another ball vertically downwards with the same speed. Which of the balls will touch the ground with higher velocity?
Answer:

The initial velocity of the 1st ball =-u and that of the 2nd ball = u. If the height of the tower is h, and v₁ and v₂ are the final velocities of balls 1 and 2 respectively just before touching the ground,

⇒ \(v_1^2=(-u)^2+2 g h=u^2+2 g h\)

⇒ \(v_2^2=(+u)^2+2 g h=u^2+2 g h\)

∴ \(v_1=v_2\)

Both the balls will touch the ground with the same velocity.

Question 22. A ball is projected vertically upwards from the ground with a velocity. After some time the ball comes back to the ground and rebounds with a velocity v₂(< v₁). Neglecting air resistance, draw the velocity-time graph for the motion of the ball.
Answer:

The required graph is given

Explanation: Let the upward velocity be positive.

t₁ = time to reach the maximum height

t₂ = time to reach maximum height after rebound.

When projected upwards with velocity v₁, the ball continues to rise up with a constant retardation due to the gravitational pull. At time t = t₁, the ball acquires a downward velocity which is taken as negative but remains under the influence of the same gravitational pull, and hence the ball returns to the ground at time t = 2t₁.

The ball rebounds with velocity v₂ and at a time t = 2t₁ + t₂, attains the maximum height. At t = 2t₁ + 2t₂, it again comes back to the ground for the second time.

Real-Life Examples of One-Dimensional Motion Problems

Example 23. State whether the displacement can be more than the total distance covered by a particle
Answer:

No, displacement cannot be more than the total distance covered by a particle. Because, to calculate the displacement, we measure the minimum, i.e., rectilinear distance between the initial and the final positions of the particle.

Example 24. Two objects are thrown vertically upwards with the same velocity v from the same point. If the second object is thrown a time T later than the first object, when will the two objects collide with each other?
Answer:

Let us assume that at a time t after throwing the first object, the two will collide with each other at a height h above the initial point.

So, for the first and the second objects,

h = \(\nu t-\frac{1}{2} g t^2 \text { and } h=\nu(t-T)-\frac{1}{2} g(t-T)^2 \)

∴ \(v t-\frac{1}{2} g t^2=\nu(t-T)-\frac{1}{2} g(t-T)^2\)

= \(\nu t-v T-\frac{1}{2} g t^2-\frac{1}{2} g T^2+g t T\)

or,  \(v T+\frac{1}{2} g T^2=g t T \quad \text { or, } t=\frac{v}{g}+\frac{T}{2} .\)

Question 25. Sketch the nature of the position-time graph for the unidirectional motion of a particle, having a variable velocity.
Answer:

The curved line OP indicates the nature of the graph. It is to be noted that, the curve OP does not have a negative gra¬dient at any point on it. The negative gradient of a line like AB means that, time is running backwards—it is physically meaningless. On the other hand, the negative gradient of CD means that the particle is moving back; then it would not be a unidirectional motion.

Question 26. Can you explain the translation of a car by the translation of a single particle? Justify your answer.
Answer:

Yes, the translation of a single particle can describe that of a car. A car is a combination of a large number of particles. By the definition of translation, each of these particles has a motion identical to that of the others. So the motion of any one of the particles is sufficient to describe the motion of the car.

Question 27. A particle travels for a time 2t₀ with velocity v = c|t- t₀|, where c is a constant. What Is the distance travelled?
Answer:

The velocities at contextual times are shown in the table:

The corresponding velocity-time graph is shown.

∴ Distance travelled in time 2 t₀

= area enclosed by the velocity-time graph and the time-axis

= area of ΔAOB + area of ΔBCD

= \(\frac{1}{2} \cdot t_0 \cdot c t_0+\frac{1}{2} \cdot\left(2 t_0-t_0\right) \cdot c t_0=\frac{1}{2} c t_0^2+\frac{1}{2} c t_0^2=c t_0^2\)

Question 28. The velocity displacement (v-x) graph of a moving particle is given. Draw the corresponding acceleration-displacement (a-x) graph.

Answer:

For the given straight-line graph, slope = \(-\frac{v_0}{x_0}\) and intercept on \(\nu-axis=v_0\)

∴ Its equation is, \(v=-\frac{v_0}{x_0} x+v_0\)

Acceleration, \(a=\frac{d v}{d t}=-\frac{v_0}{x_0} \frac{d x}{d t}=-\frac{v_0}{x_0} v=-\frac{v_0}{x_0}\left(-\frac{v_0}{x_0} x+v_0\right)\)

i.e., \(a=+\frac{v_9^2}{x_0^2} x-\frac{v_0^2}{x_0}\)

So, the required graph is a straight line of slope \(\frac{v_0^2}{x_0^2}\), with an intercept of \(-\frac{v_0^2}{x_0}\) with a-axis.

Short Notes on One-Dimensional Motion for Long Answers

Question 29. An object is thrown vertically upwards. What will be the nature of its displacement-time graph?
Answer:

The upward motion corresponds to a decreasing velocity, represented by a curve OA of decreasing slope on the x-t diagram.

On the other hand, the downward motion after attaining the highest point corresponds to an increasing velocity, represented by the curve AB on the same diagram. The curve OAB is a parabola.

Question 30. The equation x = Asinωt gives the relation between the time t and the corresponding displacement x of a moving particle, where A and ω are constants. Prove that the acceleration of the particle is proportional to its displacement and is directed opposite to it.
Answer:

x = Asinωt

∴ Velocity, v = \(\frac{dx}{dt}\) = ω A cosωt

and, acceleration, a = \(\frac{dv}{dt}\) = ωA(-ωsinωt)

= -ω²A sinωt = -ω²x

So, a ∝ -x

This means that

1. a is proportional to x,
2. The direction of a is opposite to that of x, as indicated by the negative sign.

Question 31. The velocity-time graph for a given particle is shown Draw the acceleration-time, displacement-time and distance-time graphs for the particle.

Answer:

The acceleration-time, displacement-time and distance-time graphs for the particle respectively.

Question 32. The figure below represents the acceleration-time graph of a particle at a given time. Assuming that the particle starts from rest, draw the velocity-time and displacement-time graphs for the particle.

Answer:

The velocity-time and displacement-time graphs for the particle respectively.

Question 33. Identify the types of motion: whether it is one-dimensional, two dimensional or three-dimensional,

1. Kicking a football
2. The motion of the needle clock

Answer:

1. Kicking a football produces a projectile motion. So, it is two-dimensional.
2. The motion of the needle clock is a circular Motion. So, it is two-dimensional.

Question 34. Which of the following graphs represents the one-dimensional motion of a particle? Give reasons for your answer.

Answer: All four graphs of do not represent the one-dimensional motion of a particle.

Reasons:

1. If we draw a line perpendicular to the time axis, it will cut the graph at two points which means that the particle has two different positions at the same time which is impossible. The arrows shown on the graph are meaningless.
2. If we draw a line perpendicular to the time axis, then it shows that the particle has a positive as well as a negative velocity, i.e., velocities in opposite directions at the same time, which is actually not possible in one-dimensional motion.
3. It shows that after a certain time, the total distance travelled by the particle decreases with time which is again not possible in one-dimensional motion.
4. It shows that the particle has negative speeds at certain instants. This is not a real situation because speed is always positive.

Question 35. Considering that a particle starts its motion from rest, draw the displacement-velocity graph from the given acceleration-time graph.

Answer:

The displacement-velocity graph of the given acceleration-time graph is drawn below.