Kinematics Test

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Kinematics Test - 17

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Question 1
1 / -0

Figure (i) and (ii) below show the displacement-time graphs of two particles moving along the x-axis. We can say that

A
Both the particles are having a uniformly accelerated motion

B
Both the particles are having a uniformly retarded motion

C
Particle (i) is having a uniformly accelerated motion while particle (ii) is having a uniformly retarded motion

D
Particle (i) is having a uniformly retarded motion while particle (ii) is having a uniformly accelerated motion

Solution

From equation of 2nd law of motion for uniform acceleration, we get x = x_o + 1/2 at²

Thus when acceleration or retardation is uniform, displacement time graph will be a parabola. In fig. (i) The particle is accelerated uniformly and in fig.

(ii) the particle is decelerated uniformly.
Question 2
1 / -0

For the velocity time graph shown in the figure, the distance covered by the body in the last two seconds of its motion is what fraction of the total distance travelled by it in all the seven seconds?

A
1/2

B
1/4

C
2/3

D
2/31/3

Solution

Distance covered is equal to the area under velocity-time graph.

Distance in last two seconds s² = 1/2 × 8 × 2 = 8 m

Total distance = 1/2 × 8 × (6+2) = 32 m

Fraction of the distance covered by the body in the last two seconds of its motion to total distance traveled by it in all the seven seconds is 8/32 = 1:4
Question 3
1 / -0

In the following graph, distance travelled by the body in metres is

A
200

B
250

C
300

D
400

Solution

Distance = Area covered between graph and displacement axis = 1/2 (30+10)10 = 200 meter.
Question 4
1 / -0

Velocity-time curve for a body projected vertically upwards is

A
Parabola

B
Ellipse

C
Hyperbola

D
Straight line

Solution

Because acceleration due to gravity is constant so the slope of line will be constant i.e. velocity time curve for a body projected vertically upwards is straight line.
Question 5
1 / -0

The displacement-time graph of moving particle is shown below

The instantaneous velocity of the particle is negative at the point

A
D

B
F

C
C

D
E

Solution

Slope of dispacement time graph is nagative only at point E.
Question 6
1 / -0

An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) – velocity (v) graph of this object is

A

B

C

D

Solution

i.e. graph should be parabola symmetric to displacement axis.
Question 7
1 / -0

Which of the following graph represents uniform motion

A

B

C

D

Solution

This graph shows uniform motion because line having a constant slope.
Question 8
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A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the ground is

A

B

C

D

Solution
Question 9
1 / -0

The graph of displacement v/s time is

Its corresponding velocity-time graph will be

A

B

C

D

Solution

We know that the velocity of body is given by the slope of displacement – time graph. So it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of graph) and then it will becomes negative.
Question 10
1 / -0

A train moves from one station to another in 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is

A
140 km h^–2

B
160 km h^–2

C
100 km h^–2

D
120 km h^–2

Solution

Maximum acceleration will be represented by CD part of the graph