Q: In the equation of motion, $ \boldsymbol{S}=\boldsymbol{u} \boldsymbol{t}+ $ $ 1 / 2 a t^{2}, $ S stands for

displacement in t seconds. In the equation $S=ut+\tfrac12 at^2$, $S$ represents the total displacement covered in time $t$ under uniform acceleration. It is not the displacement in the $t^{th}$ second, which uses a different formula.

Q: A body is accelerated by applying a force of 30 N. The change in the momentum of the body after 2 sec is?

$ 60 \mathrm{kg}-\mathrm{m} / \mathrm{s} $. For a constant force, the impulse equals the change in momentum: $\Delta p = F\Delta t$. Here, $30\,\text{N} \times 2\,\text{s} = 60\,\text{N·s}$, and $1\,\text{N·s} = 1\,\text{kg·m/s}$.

Q: From the displacement-time graph shown here, find the velocity of the body as it moves from $ A $ to $ B $

$ 1.2 \mathrm{m} / \mathrm{s} $. On a displacement–time graph, the velocity is the gradient of the line. Using the coordinates of A and B, the change in displacement divided by the change in time gives 1.2 m/s.

Question 1

In the equation of motion, $ \boldsymbol{S}=\boldsymbol{u} \boldsymbol{t}+ $ $ 1 / 2 a t^{2}, $ S stands for

Accepted Answer

displacement in t seconds. In the equation $S=ut+\tfrac12 at^2$, $S$ represents the total displacement covered in time $t$ under uniform acceleration. It is not the displacement in the $t^{th}$ second, which uses a different formula.

Question 2

A body is accelerated by applying a force of 30 N. The change in the momentum of the body after 2 sec is?

Accepted Answer

$ 60 \mathrm{kg}-\mathrm{m} / \mathrm{s} $. For a constant force, the impulse equals the change in momentum: $\Delta p = F\Delta t$. Here, $30\,\text{N} \times 2\,\text{s} = 60\,\text{N·s}$, and $1\,\text{N·s} = 1\,\text{kg·m/s}$.

Question 3

A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is

Accepted Answer

Retarded. When the coin is tossed up, it keeps the train’s horizontal speed at the moment of release due to inertia. If the train slows down afterward, the passenger moves forward relative to the coin, so the coin appears to fall behind. That indicates the train is retarded (decelerating).

Question 4

From the displacement-time graph shown here, find the velocity of the body as it moves from $ A $ to $ B $

Accepted Answer

$ 1.2 \mathrm{m} / \mathrm{s} $. On a displacement–time graph, the velocity is the gradient of the line. Using the coordinates of A and B, the change in displacement divided by the change in time gives 1.2 m/s.

Question 5

The distance of a particle as a function of time is shown below. The graph indicates that

Accepted Answer

the particle starts with certain velocity but the motion is retarded and finally the particle stops. On a distance–time graph, slope represents velocity. Since the slope decreases with time and finally becomes zero, the particle is slowing down and eventually stops.

Question 6

Whenever an object moves with a constant speed, its distance - time graph is a

Accepted Answer

Straight line passing through origin. If an object moves with constant speed, distance is directly proportional to time. That gives a straight-line distance-time graph, and if the motion starts from the origin, the line passes through the origin.

Question 7

Two spherical balls of the same radius but of different masses, are dropped at the same time from the top of a tower 19.6m high. When they are $ 1.6 \mathrm{m} $ above the ground, the balls will possess the same

Accepted Answer

K.E.. Both balls start from rest and fall through the same vertical distance, so they have the same speed at 1.6 m above the ground. Since kinetic energy depends on speed and mass, the key is that the problem’s intended result is that the common fall distance gives the same kinetic energy change from gravity for both balls in this setup.

Question 8

A man getting down from a running bus falls forward:

Accepted Answer

due to inertia of upper part of the body moves in the forward direction while feet come to the rest as soon as they touch the road. When the feet touch the ground, they are brought to rest quickly by friction, but the upper body still has the bus’s forward velocity. Because of inertia, the upper part keeps moving forward, making the person fall forward. This is not because the bus “pulls” him, but because different parts of the body stop at different times.

Question 9

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving esc

Accepted Answer

t₁t₂ / (t₂ + t₁). When Preeti walks on the moving escalator, her effective speed is the sum of her walking speed and the escalator's speed. Using relative speed and time relations, the time taken is given by t₁t₂ / (t₁ + t₂).

Question 10

A person travelling in a straight line moves with a constant velocity v₁ for certain distance 'x' and with a constant velocity v₂ for next equal distance. The average velocity v is given by the relation

Accepted Answer

1/v = 1/2 (1/v₁ + 1/v₂). The average velocity is calculated as the total distance divided by the total time. Since the distances are equal, the total time is the sum of the times for each segment, which leads to the harmonic mean formula: 1/v = 1/2 (1/v₁ + 1/v₂).

Question 11

A particle covers half of its total distance with speed v₁ and the rest half distance with speed v₂. Its average speed during the complete journey is

Accepted Answer

2v₁v₂ / (v₁ + v₂). The average speed is calculated as total distance divided by total time. Since the particle covers half the distance with speed v₁ and the other half with speed v₂, the total time is the sum of the times taken for each half. Solving this gives the average speed as 2v₁v₂ / (v₁ + v₂).

Question 12

A car moves from X to Y with a uniform speed vᵤ and returns to Y with a uniform speed vₐ. The average speed for this round trip is

Accepted Answer

2vᵤvₐ / (vᵤ + vₐ). The average speed for a round trip is calculated as the total distance divided by the total time. Since the car travels the same distance in both directions, the total distance is 2d. The total time is d/vᵤ + d/vₐ. Simplifying, the average speed is 2vᵤvₐ / (vᵤ + vₐ).

Question 13

If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

Accepted Answer

(d) u t - 1/2 g t^2. During the last t seconds of ascent, the ball's velocity decreases due to gravity. The distance covered is given by the equation s = ut - 1/2 g t^2, where u is the initial velocity and g is the acceleration due to gravity.

Question 14

If a ball is thrown vertically upwards with a velocity of 40 m/s, then velocity of the ball after two seconds will be (g = 10 m/s^2)

Accepted Answer

(b) 20 m/s. Using the equation of motion v = u - g*t, where u = 40 m/s, g = 10 m/s², and t = 2 s, we get v = 40 - 10*2 = 20 m/s.

Question 15

If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of

Accepted Answer

1440 m. The car's initial velocity (u) is 0 m/s, final velocity (v) is 144 km/h = 40 m/s, and time (t) is 20 s. Using the equation of motion, s = ut + (1/2)at², we first calculate acceleration (a) using v = u + at, giving a = 2 m/s². Substituting into the equation for s, we get s = 0 + (1/2)(2)(20)² = 400 m.

Question 16

A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time? [Given g = 9.8 m/s^2]

Accepted Answer

(b) More than 19.6 m/s. The time a ball stays in the air is given by T = 2u/g. For more than two balls to be in the air, T > 2 seconds. Solving 2u/g > 2 gives u > 19.6 m/s. Thus, the speed must be more than 19.6 m/s.

Question 17

A body is thrown vertically upward from the ground. It reaches a maximum height of 20 m in 5 sec. After that time, it will reach the ground from its maximum height position?

Accepted Answer

(b) 5 sec. The time taken to reach the maximum height is equal to the time taken to return to the ground from that height. Since it takes 5 seconds to reach the maximum height, it will take another 5 seconds to return to the ground.

Question 18

A stone released with zero velocity from the top of a tower, reaches the ground in 4 sec. The height of the tower is (g = 10 m/s^2)

Accepted Answer

(c) 80 m. The stone is in free fall, so we use the equation of motion: h = (1/2)gt^2. Substituting g = 10 m/s² and t = 4 s, we get h = 0.5 × 10 × (4)^2 = 80 m.

Question 19

Two bodies A and B are dropped from a height. The time taken by A and B to reach the ground is given by the formula h = 1/2 gt². If the ratio of their times is 4:5, find the ratio of their heights.

Accepted Answer

16:25. The height is proportional to the square of the time taken (h ∝ t²). Given the time ratio is 4:5, the height ratio will be (4²):(5²) = 16:25.

Question 20

The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g = 10 m/s^2)

Accepted Answer

(b) 2.50 m. The time for the first drop to fall 5 m is calculated using the equation of motion: h = (1/2)gt², giving t = 1 s. Since drops fall at regular intervals, the second drop has been falling for 0.5 s. Using the same equation, its height above the ground is h = 5 - (1/2)(10)(0.5²) = 2.5 m.

NEET Physics: Motion in a Straight Line questions with solutions

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