Kinematics - NEET Physics Questions
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Kinematics

Practice Questions:
Question 121: easy

A particle begins to move along straight line where the acceleration \( (a) \) of the particle varies with displacement \( (x) \) according to relation, \( a = 5x \), then velocity of the particle varies with displacement as

1. \( x^{1/2} \)
2. \( x^1 \)
3. \( x^{1/3} \)
4. \( x^{3/4} \)
View Answer

Using \( a = v \frac{dv}{dx} \), we write \( v \frac{dv}{dx} = 5x \). Integrating both sides, \( \int v \, dv = \int 5x \, dx ⇒ \frac{v^2}{2} = \frac{5x^2}{2} + C \). Assuming the particle starts from rest, \( v^2 \propto x^2 ⇒ v \propto x^1 \).

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Question 122: easy

A particle is projected with a speed of \( 20 \, \text{m s}^{-1} \) from level ground at an angle \( \theta \) equal to \( 45^\circ \) from horizontal. The ratio of maximum height attained by the body to horizontal range acquired by the body will be

1. \( \frac{2}{1} \)
2. \( \frac{1}{2} \)
3. \( \frac{1}{4} \)
4. \( \frac{3}{2} \)
View Answer

The ratio of maximum height \( H \) to horizontal range \( R \) is given by \( \frac{H}{R} = \frac{\tan \theta}{4} \). For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \), leading to \( \frac{H}{R} = \frac{1}{4} \).

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Question 123: easy

A car starts from rest, accelerates uniformly at \( 2 \, \text{m/s}^2 \). The distance travelled by the car in fourth second is

1. 2.25 m
2. 14 m
3. 7 m
4. Zero
View Answer

The distance travelled in the \( n^{\text{th}} \) second is \( s_n = u + \frac{a}{2}(2n - 1) \). Substituting \( u = 0 \), \( a = 2 \, \text{m/s}^2 \), and \( n = 4 \) yields \( s_4 = 0 + \frac{2}{2}(2(4) - 1) = 7 \, \text{m} \).

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Question 124: easy

A particle moves along a straight line with velocity given by \( v = (6 – 3t) \) where \( v \) is in \( \text{m/s} \) and \( t \) in seconds. Determine when the particle returns to its starting point.

1. \( 4\text{ s} \)
2. \( 2\text{ s} \)
3. \( 3\text{ s} \)
4. \( 5\text{ s} \)
View Answer

Displacement is \( S = \int v \, dt = \int_0^t (6 - 3t) \, dt = 6t - 1.5t^2 \). Returning to the starting point means \( S = 0 ⇒ 6t - 1.5t^2 = 0 ⇒ t = 4\text{ s} \).

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Question 125: 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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Question 126: easy

Two particles A and B are projected from ground at an angle of \( 30^\circ \) with the horizontal with velocity \( 20\text{ m/s} \) and \( 40\text{ m/s} \) respectively. The maximum height and time of flight are both greater for which particle?

1. A
2. B
3. Same for both
4. Data insufficient
View Answer

Maximum height \( H \propto u^2 \) and Time of flight \( T \propto u \) for a given angle. Since particle B has a larger initial velocity, both parameters are greater for B.

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