Kinematics

Practice Questions:

Question 1: easy

A car moves from X to Y with a uniform speed \[v_{u}\] and returns to X with a uniform speed \[v_{d}\]. The average speed for this round trip is

1. \[\frac{2v_{d}v_{u}}{v_{d}+v_{u}}\]

2. \[\sqrt{v_{d}v_{u}}\]

3. \[\frac{v_{d}v_{u}}{v_{d}+v_{u}}\]

4. \[\frac{v_{u}+v_{d}}{2}\]

View Answer

\[ \frac{2}{V_{av}}= \frac{1}{V_{1}}+\frac{1}{V_{2}} \]

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Question 2: easy

The velocity of a body of mass 20 kg decreases from 20 ms–¹ to 5 ms–¹ in a distance of 100 m. Force on the body is

1. –27.5 N

2. –47.5 N

3. –37.5 N

4. –67.5 N

View Answer

We can use the work-energy principle to find the force.

Given:
- Initial velocity, \( u = 20 \, \text{m/s} \)
- Final velocity, \( v = 5 \, \text{m/s} \)
- Distance, \( s = 100 \, \text{m} \)
- Mass, \( m = 20 \, \text{kg} \)

Using the equation of motion:
\[
v^2 = u^2 + 2as
\]
\[
(5)^2 = (20)^2 + 2a(100)
\]
\[
25 = 400 + 200a
\]
\[
200a = -375 ;a = -\frac{375}{200} = -1.875 \, \text{m/s}^2
\]

Now, force \( F = ma \):
\[
F = 20 \times (-1.875) = -37.5 \, \text{N}
\]

Thus, the force on the body is -37.5 N (opposite to the direction of motion).

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Question 3: easy

The displacement-time graph of two moving particles A and B respectively make angles of 30° and 45° with the x-axis respectively. The ratio of their velocities is

1. 1 : √3

2. 1 : 2

3. 1 : 1

4. √3 : 2

View Answer

Slope of displacement time graph represents velocity.

V_A= tan 30 ⁰
V_B= tan 45 ⁰

So, V_A/V_B=tan 30 ⁰/tan 45 ⁰= 1/√3

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Question 4: easy

A body is travelling in a straight line with a uniformly increasing speed. Which one of the plot represents the change in distance (s) travelled with time (t)?

View Answer

\[ s= \frac{1}{2}at^{2}\]

As s is proportional to square of t. s-t graph is a parabola.

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Question 5: easy

A particle is thrown upwards, then correct v-t graph will be

View Answer

Slope of v-t graph represent acceleration. when object is thrown upward its acceleration is -g for the entire journey. So, slope is -g.

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Question 6: easy

A body starts from rest and moves with uniform acceleration. Which of the following graphs represents it motion?

View Answer

\[ v = u + at ; as, u = 0 ; v = at\]

As, v is proportional to t ; graph of v-t is a straight line.

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Question 7: easy

If the figure below represents a parabola, identify the physical quantities representing Y and X for constant acceleration

1. X = time, Y = velocity

2. X = velocity, Y = time

3. X = time, Y = displacement

4. X = time, Y = acceleration

View Answer

\[ S= \frac{1}{2}at^{2}\]

As, S is proportional to t graph is a parabola

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Question 8: easy

Two bodies of different masses \[m_{a} and m_{b}\] are dropped from two different heights, viz, a and b. The ratio of time taken by the two to drop through these distances is :

1. a : b

2. \[\frac{m_{a}}{m_{b}}:\frac{b}{a}\]

3. √a : √b

4. a² : b²

View Answer

The time taken by an object to fall from a height is given by the equation:

\[
t = \sqrt{\frac{2h}{g}}
\]

For mass \( m_a \) dropped from height \( a \), the time taken is:

\[
t_a = \sqrt{\frac{2a}{g}}
\]

For mass \( m_b \) dropped from height \( b \), the time taken is:

\[
t_b = \sqrt{\frac{2b}{g}}
\]

The ratio of time taken by the two bodies is:

\[
\frac{t_a}{t_b} = \frac{\sqrt{\frac{2a}{g}}}{\sqrt{\frac{2b}{g}}} = \sqrt{\frac{a}{b}}
\]

So, the ratio of the time taken is:

\[
\frac{t_a}{t_b} = \sqrt{\frac{a}{b}}
\]

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Question 9: easy

A particle is thrown vertically upwards with a velocity ν. It returns to the ground in time T. Which of the following graphs correctly represents the motion?