Motion Test

Result

Motion Test - 3

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Question 1
1 / -0

Let the velocity and acceleration of a body be denoted by v and a, respectively. Which of the following relations holds true?

A
a can be non-zero, when v = 0.

B
a must be zero, when v = 0.

C
a > 0, when v > 0.

D
a < 0, when v > 0.

Solution

Acceleration of a body may be non-zero even if the velocity is zero. For example, if we throw a ball upwards, acceleration due to gravity is acting in the downward direction; at the highest point, the velocity of the ball becomes zero but acceleration is acting downwards.
The object will speed up if the direction of the velocity and that of the acceleration are the same, and will slow down if both are opposite in direction. There is no definite relationship between acceleration and velocity.
Question 2
1 / -0

The displacement-time graphs for two particles A and B are straight lines inclined at angles of 30° and 60° with the time axis, respectively. What is the ratio of the speeds v_A and v_B, respectively?

A
1 : 2

B
1 :

C
: 1

D
1 : 3

Solution

Slope of the displacement-time graph gives the velocity.
Slope of the graph = tan θ
Question 3
1 / -0

A tennis ball is dropped so that it falls vertically on the floor and bounces again. Taking velocity upwards as positive, which of the following graphs best represents the variation of its velocity v with time t?

A

B

C

D

Solution

As the ball is dropped from a certain height 'h', it is moving downwards with increasing speed as the acceleration due to gravity is in downward direction (v = u - gt). So, v-t graph will be a straight line. At the time of impact (which is very small as compared to the time of travel), the direction of the velocity will change to upward but the acceleration is acting in downward direction, so the magnitude of the velocity will decrease.
Question 4
1 / -0

A car is moving along a straight road with uniform acceleration. It passes through two points P and Q with velocities 30 km/hr and 40 km/hr, respectively. What is the velocity of the car mid-way between P and Q?

A
33.3 km/hr

B
20km/hr

C
km/hr

D
35 km/hr

Solution

As the object is moving with constant acceleration, we can use:

Let R be the point mid-way between P and Q, and let the velocity of the car at point R be v.
PR = RQ = s
For the distance PR, we have ---1
For the distance RQ, we have
---2

From 1 and 2, we havekm/hr
Or km/hr
Question 5
1 / -0

A ball is thrown upwards with a velocity v. It attains a height of 50 m and comes back to the thrower. Which of the following statements is correct?

A
The total distance covered is zero.

B
The total distance covered is 50 m.

C
The displacement is 104 m.

D
The time taken is 2s.

Solution

As the ball reaches the hands of the thrower, the displacement of the ball is zero.
Distance travelled during upward journey = 50 m
Total distance travelled during upward and downward journey = 50 + 50 = 100 m

Time taken by the ball to reach the hands of thrower,
Question 6
1 / -0

With what speed should a body be thrown upwards so that the distances travelled in the 5^thsecond and 6^th second are equal?

A
m/s

B
49 m/s

C
58.4 m/s

D
98 m/s

Solution

As the distances travelled in the 5^th second and the 6^th second are the same, so we can conclude that the particle reaches the highest point at t = 5 s.
Time taken to reach the highest point is 5 s.
At the highest point, v = 0 (a = -g)
Using
m/s
Question 7
1 / -0

A ball is released from the top of height h metres. It takes T seconds to reach the ground. Where is the ball at T/2 second?

A
At (h/4) m from the ground

B
At (h/2) m from the ground

C
At (3h/4) m from the ground

D
Depends upon the mass and volume Depends upon the mass and volume

Solution

As ,
u = 0
Let 'h' be the height of the tower.
'h' is the distance travelled in time T/2.
Then,

Hence, distance of the ball from the ground =
Question 8
1 / -0

Water droplets fall at regular intervals from a roof. At an instant, when a droplet is about to fall, what will be the ratio of separation between 3 successive droplets below the roof?

A
1 : 2 : 3

B
1 : 4 : 9

C
1 : 3 : 5

D
1 : 5 : 13

Solution

Suppose the droplets fall at an interval of 1 sec.
Distance travelled by the object in n^th second, As u = 0

Hence, the ratio of separation between three successive droplets will be as given below:
Question 9
1 / -0

A person has covered a certain distance with average velocity of 20 m/s. First 1/3^rdof the total distance is covered within 2/3^rdof the total time with velocity v₁ and the rest of the distance is covered with velocity v₂ in 1/3^rdof the total time, then v₁and v₂ are (in m/s)

A
20 and 30

B
5 and 20

C
10 and 20

D
10 and 40

Solution

As the average velocity is 20 m/s,

Let the distance travelled be 'd' and time taken be 'T'.

For the first 1/3^rd distance,

For the first 2/3^rd distance,
Question 10
1 / -0

A car is moving in a straight line whose velocity-time graph is given.

What is the displacement of the car in first 10 seconds?

A
30 m

B
32 m

C
34 m

D
36 m

Solution

Area above the time axis is taken as positive and below the time axis, it is taken as negative. Velocity at t = 10 s is -2 m/s, negative sign indicates that the direction of the velocity is opposite to the initial direction.

As the slope of the graph is -1 m/s²,
Area above the time axis,

Area below the time axis,

Displacement of the car = A₁ + A₂ = 32 + (-2) = 30 m
Question 11
1 / -0

An object is moving along the straight line with acceleration along negative x-axis, then

A
it is definitely moving with increasing velocity

B
it is definitely moving with decreasing velocity

C
it may be moving with increasing velocity

D
it is definitely moving with constant acceleration

Solution

As the motion is accelerated, it may be constant or time dependent.
If the velocity of the object is along the positive x-axis and the direction of the acceleration is along the negative x-axis, then the motion is retarded.
If the velocity of the object is along the negative x- axis and the direction of the acceleration is along negative x-axis, then the motion is accelerated.
Question 12
1 / -0

A balloon initially at rest, starts moving upwards with a constant acceleration of 2 m/s². A ball is dropped from the balloon after 10 seconds of flight. Maximum height attained by the ball during the motion is (g = 10 m/s²)

A
20 m

B
80 m

C
100 m

D
120 m

Solution

Take upward direction as positive. As u = 0 m/s,
Using
Height attained by the balloon in 10 seconds,
Velocity attained by the balloon in 10 seconds,

Hence, the velocity of the ball with respect to earth is 20 m/s.
At the maximum height, the velocity of the ball becomes zero.
Using v² - u² = 2as_,

Hence, maximum height attained by the ball, H = 100 + 20 = 120 m
Question 13
1 / -0

A small ball is dropped from the balloon moving vertically upwards at a speed of 10 m/s when the balloon is at a height of 15 m from the ground. Neglect air friction and take g = 10 m/s². Which of the following is suitable for the present situation?

A
The ball reaches the ground in 3 s.

B
The ball covers a distance of 20 m.

C
The magnitude of the average velocity of the ball is 8.33 m/s.

D
The ball moves upwards at a speed of 5 m/s at the instant when it is dropped from the balloon.

Solution

As the ball was moving with the balloon, so at the instant when it was dropped from the balloon, its velocity with respect to the balloon was zero, but with respect to earth it was moving upwards with the same speed as that of the balloon. Taking upward direction as positive and the ground as origin,
Initial velocity of the ball, u = 10 m/s
Initial position of the ball, y₀= 15 m, final position y = 0.

Using

We have

Hence the ball will reach the ground in 3 seconds.

Let s be the distance travelled by the ball before coming to rest for a moment at the highest point.
Using

Hence, total distance covered by the ball before striking the ground = 5 m upward + 5 m downward + 15 m downward = 25 m
Question 14
1 / -0

A launch takes 3 hrs to go downstream from point A to B and 6 hrs to come back to A from B. The time taken by the launch to cover the same distance downstream when its engine cut-off is

A
4.5 hrs

B
9 hrs

C
12 hrs

D
18 hrs

Solution

Let v be the velocity of the launch and v_r be the velocity of the stream.

During downstream, speed of the launch, v_d = v + v_r
---1
Here, t₁ is the time taken from A to B. During upstream, speed of the launch, v_r = v - v_r
---2
Here, t₂ is the time taken from B to A.
Let the time taken by the launch when the engine is cut-off from A to B be t.
Then,

From 1 and 2,

Alternate method:
Let the distance be 1 km.
During downstream, speed of the launch,
- 1
During upstream, speed of the launch,
-2
From 1 and 2,

Speed of the launch with cut-off engine,
Hence, time taken is t = 12 hrs
Question 15
1 / -0

A car driver travelling with a uniform velocity of 2 m/s notices a railway level crossing at a distance of 435 m from him.And also he notices that it is going to be closed in 10 seconds. First he decides to cross the level crossing; hence, he accelerates his car at the rate of 2 ms^-² for 5 seconds. Then he decides to stop the car. So he applies brake and stops the car exactly before the level crossing (without following the timer). Calculate the minimum rate at which he has to decelerate the car so that he stops the car exactly before the level crossing.

A
0.18 m/s²

B
1.8 m/s²

C
3.6 m/s²

D
18 m/s²

Solution

Velocity after 5 seconds,

Distance travelled in 5 seconds,

Distance to be covered before crossing = 435
- 35 = 400 m
Final velocity = 0 m/s

So,the retardation of car is 0.18 m/s².
Question 16
1 / -0

If the area under given velocity-time graph can be changed into acceleration-time graph, then which one of the given options represents acceleration-time graph:

A

B

C

D

Solution

The slope of the velocity-time graph gives the acceleration.
If the slope is constant, acceleration is constant; if the slope is positive, acceleration is positive; and if the slope is negative, acceleration is negative.
Question 17
1 / -0

A balloon is moving up from the ground in such a way that its acceleration is linearly decreasing with its height above the ground. It starts from the ground with an acceleration of 4 m/s² and with zero initial velocity. Its acceleration becomes zero at a height of 3 m. The speed of the balloon at a height of 1.5 m is

A
4 m/s

B
8 m/s

C
6 m/s

D
3 m/s

Solution

For an object moving with constant acceleration, we have

Here, the product 'as' represents the area under the acceleration-position graph.
As per the question, the acceleration decreases linearly with the height, so we can say that at the height of 1.5 m, the acceleration is 2 m/s².

Area of the shaded portion (trapezium) =
Hence, we have
(as u = 0)
v = 3 m/s
Question 18
1 / -0

What is the ratio of the average speed for the first two seconds to the average speed for the next four seconds?

A
1 : 1

B
3 : 1

C
2 : 1

D
1 : 2

Solution

v =

As shown in the given figure:

Therefore, the ratio of speeds

v₁ : v₂ = 2 : 1
Question 19
1 / -0

A body is moving in a circle, as shown in the figure, starting from point A in the clockwise direction.

It completes one cycle in 20 sec.

What will be the average velocity as it starts from A and reaches C after completing two cycles?

A
m/s towards northeast

B
m/s towards southeast

C
m/s towards southeast

D
m/s towards southwest

Solution

Time taken in 2 cycles = 20 + 20 = 40 sec
Time taken to cover the distance from A to C in clockwise direction
= th of the time taken to complete 1 cycle
= × 20 = 15 sec
Total time = 40 + 15 = 55 sec
Displacement = AC = = 5
Average velocity = m/s in South-East direction.
Question 20
1 / -0

If a particle is thrown above, then the correct v-t graph will be

A

B

C

D

Solution

Take initial position as the origin and direction of motion (i.e. vertically up) as positive. As the particle is thrown with an initial velocity, therefore at the highest point, its velocity will be zero and then it returns back to its reference position. This situation is best depicted in figure of option (a).

In the figure above, part AB denotes upward motion and part BC denotes downward motion.