Kinematics Test

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Kinematics Test - 16

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Question 1
1 / -0

The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in four seconds is

A
60 m

B
55 m

C
25 m

D
30 m

Solution
Question 2
1 / -0

The displacement of a particle as a function of time is shown in the figure. The figure shows that

A
The particle starts with certain velocity but the motion is retarded and finally the particle stops

B
The velocity of the particle is constant throughout

C
The acceleration of the particle is constant throughout.

D
The particle starts with constant velocity, then motion is accelerated and finally the particle moves with another constant velocity

Solution

The slope of displacement-time graph goes on decreasing, it means the velocity is decreasing i.e. It's motion is retarded and finally slope becomes zero ie. particle stops.
Question 3
1 / -0

A ball is thrown vertically upwards. Which of the following graph/graphs represent velocity-time graph of the ball during its flight (air resistance is neglected)

A

B

C

D

Solution

In the positive region the velocity decreases linearly (during rise) and in the negative region velocity increases linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in the negative region.
Question 4
1 / -0

The graph between the displacement x and time t for a particle moving in a straight line as shown in figure. During the interval OA, AB, BC and CD, the acceleration of the particle is :

A
+ 0 + +

B
- 0 + 0

C
+ 0 - +

D
- 0 - 0

Solution

Slope of the displacement-time graph gives the instantaneous velocity of the particle.

Region OA : Slope is (+) and decreasing i.e   v decreases in positive direction.

⟹ The particle is retarding i.e   a  is negative (-)

Region AB : Slope is zero i.e   v=constant .

⟹ The particle is moving with zero acceleration i.e   a=0

Region BC : Slope is (+) and increasing i.e   v increases in positive direction.

⟹ The particle is accelerating i.e   a  is positive (+)

Region CD : Slope is positive but constant i.e   v=constant .

⟹ The particle is moving with zero acceleration i.e   a=0
Question 5
1 / -0

The v-t graph of a moving object is given in figure. The maximum acceleration is

A
1 cm /sec²

B
2 cm / sec²

C
3 cm / sec²

D
6 cm / sec²

Solution

Maximum acceleration means maximum change in velocity in minimum time interval.
Question 6
1 / -0

The displacement versus time graph for a body moving in a straight line is shown in figure. Which of the following regions represents the motion when no force is acting on the body

A
ab

B
bc

C
cd

D
de

Solution

In part cd displacement-time graph shows constant slope e. velocity is constant. It means no acceleration or no force is acting on the body.
Question 7
1 / -0

The x–t graph shown in figure represents

A
Constant velocity

B
velocity of the body is continuously changing

C
Instantaneous velocity

D
The body travel with constant speed upto time t₁ and then stops

Solution

Up to time t_₁ slope of the graph is constant and after t_₁ slope is zero i.e. the body travel with constant speed up to time t_₁and then stops.
Question 8
1 / -0

A lift is going up. The variation in the speed of the lift is as given in the graph. What is the height to which the lift takes the passengers

A
3.6 m

B
28.8 m

C
36.0 m

D
Cannot be calculated from the above graph

Solution

Area of trapezium = 1/2 × 3.6 × (12+8) = 36.0 m
Question 9
1 / -0

The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 sec are respectively

A
8 m, 16 m

B
16 m, 8 m

C
16 m, 16 m

D
8 m, 8 m

Solution
Question 10
1 / -0

Velocity-time (v-t) graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is

A
60 m

B
50 m

C
30 m

D
40 m

Solution

Between time interval 20 sec to 40 sec, there is non-zero acceleration and retardation. Hence distance travelled during this interval

= Area between time interval 20 sec to 40 sec

= 1/2 × 20 × 3 + 20 × 1 = 30 + 20 = 50 m.