Relative Motion in One Dimension

Practice Questions:

Question 21: moderate

When a motorcycle moving with a uniform speed \(11\text{ m/s}\) is at a distance \(24\text{ m}\) from a car, the car starts from rest and moves with a uniform acceleration \(2\text{ m/s}^2\) away from the motorcycle. If the car begins motion at \(t = 0\), time at which the motorcycle will overtake the car is \(t = \):

1. \(8\text{ sec}\)

2. \(6\text{ sec}\)

3. \(3\text{ sec}\)

4. \(1.5\text{ sec}\)

View Answer

Distance equation for meeting: \(11t = 24 + \frac{1}{2}(2)t^2 \Rightarrow t^2 - 11t + 24 = 0\). Solving this quadratic equation gives \(t = 3\text{ s}\) and \(t = 8\text{ s}\). The first overtake occurs at \(t = 3\text{ s}\).

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Question 22: easy

A car is moving with velocity of 20 m/s on a straight road. A scooterist wishes to overtake the car in 60 s. If the car is at a distance of 1.5 km ahead, then the velocity with which the scooterist has to chase the car is

1. 25 m/s

2. 20 m/s

3. 45 m/s

4. 50 m/s

View Answer

Relative velocity required to cover 1500 m in 60 s is \(v_{\text{rel}} = \frac{1500}{60} = 25\text{ m/s}\). Since \(v_{\text{rel}} = v_s - v_c ⇒ v_s = v_c + v_{\text{rel}} = 20 + 25 = 45\text{ m/s}\).

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Question 23: easy

Assertion (A): If separation between two particles does not change then their relative velocity will be zero.

Reason (R): Relative velocity is the rate of change of position of one particle with respect to another.

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

Assertion (A): If the separation \(\vec{r}_{rel}\) is constant, it implies \(\vec{r}_{rel} \cdot \vec{v}_{rel} = 0\), meaning \(\vec{v}_{rel}\) is perpendicular to \(\vec{r}_{rel}\), but not necessarily zero (e.g., two particles orbiting each other at constant distance). So (A) is False.

Reason (R): Relative velocity is defined as the time derivative of the relative position vector. So (R) is True.

Since (A) is false and (R) is true, none of the given options are strictly correct. However, if (A) is false, options (1), (2), (3) are ruled out, leaving (4) by elimination, despite (R) being true.

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Question 24: easy

Assertion (A): The magnitude of velocity of A with respect to B will be always less than (V_A).

Reason (R): The velocity of A with respect to B is given by \(\vec{V}_{AB} = \vec{V}_A – \vec{V}_B\).

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (3) (A) is true but (R) is false

4. (4) Both (A) and (R) are false

View Answer

Assertion (A): The relative velocity is \(\vec{V}_{AB} = \vec{V}_A - \vec{V}_B\). If \(\vec{V}_B)\) is in the opposite direction to \(vec{V}_A\), then \(|vec{V}_{AB}| = |vec{V}_A| + |vec{V}_B|\), which is greater than (\|vec{V}_A|\). Thus, (A) is False.

Reason (R): The definition of relative velocity of A with respect to B is \(\vec{V}_{AB} = \vec{V}_A - \vec{V}_B\). So (R) is True.

Since (A) is false and (R) is true, none of the given options are strictly correct. However, if (A) is false, options (1), (2), (3) are ruled out, leaving (4) by elimination, despite (R) being true.

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Question 25: easy

A car is moving with velocity of \( 20\text{ m/s} \) on a straight road. A scooterist wishes to overtake the car in \( 60\text{ s} \). If the car is at a distance of \( 1.5\text{ km} \) ahead, then the velocity with which the scooterist has to chase the car is

1. \( 25\text{ m/s} \)

2. \( 20\text{ m/s} \)

3. \( 45\text{ m/s} \)

4. \( 50\text{ m/s} \)

View Answer

Relative velocity required: \( v_{\text{rel}} = \frac{\text{distance}}{\text{time}} = \frac{1500\text{ m}}{60\text{ s}} = 25\text{ m/s} \). Since \( v_{\text{rel}} = v_s - v_c \), we get \( v_s = v_c + v_{\text{rel}} = 20 + 25 = 45\text{ m/s} \).

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