Kinematics - NEET Physics Questions
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Kinematics

Practice Questions:
Question 41: easy

A train started from rest from a station and accelerated at \(2\text{ ms}^{-2}\) for 10 s. Then, it ran at constant speed for 30 s and thereafter it decelerated at \(4\text{ ms}^{-2}\) until it stopped at the next station. The distance between two station is

1. 650 m
2. 700 m
3. 750 m
4. 800 m
View Answer

Max velocity \( v = 20 \text{ ms}^{-1}\). Acceleration distance \(s_1 = \frac{1}{2}(2)(10^2) = 100\text{ m}\). Constant velocity distance \(s_2 = 20 \times 30 = 600\text{ m}\). Deceleration distance \(s_3 = \frac{20^2}{2 \times 4} = 50\text{ m}\).

Total distance = 750 m.

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Question 42: easy

A particle moves along a straight line such that its displacement at any time t is given by \(s = t^3 – 6t^2 + 3t + 4\). The velocity when its acceleration is zero is

1. \(2\text{ ms}^{-1}\)
2. \(12\text{ ms}^{-1}\)
3. \(-9\text{ ms}^{-1}\)
4. \(4\text{ ms}^{-1}\)
View Answer

Velocity \(v = 3t^2 - 12t + 3\) and acceleration \(a = 6t - 12\). Acceleration is zero at t = 2 s. Substituting \(t = 2\) in velocity equation gives \(v = 3(4) - 24 + 3 = -9 \text{ ms}^{-1}\).

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Question 43: easy

The position of a particle is given by \(\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}\) where \(t\) is in seconds and \(r\) in meter. Find the magnitude and direction of velocity \(v(t)\), at \(t = 1\text{ s}\), with respect to x-axis.

1. \(3\sqrt{2}\text{ ms}^{-1}, 30^\circ\)
2. \(3\sqrt{2}\text{ ms}^{-1}, 45^\circ\)
3. \(4\sqrt{2}\text{ ms}^{-1}, 45^\circ\)
4. \(4\sqrt{2}\text{ ms}^{-1}, 60^\circ\)
View Answer

Velocity is \(\vec{v} = \frac{d\vec{r}}{dt} = 4\hat{i} + 4t\hat{j}\). At \(t = 1\text{ s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Thus, magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2}\text{ ms}^{-1}\) and angle \(\tan\theta = \frac{4}{4} = 1 ⇒ \theta = 45^\circ\) with respect to the x-axis.

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Question 44: easy

The displacement \(s\) of a point moving in a straight line is given by : \(s = 8t^2 + 3t – 5\) where \(s\) is in cm and \(t\) in s. The initial velocity of the particle is :

1. 3 cm/s
2. 16 cm/s
3. 19 cm/s
4. Zero
View Answer

Velocity is given by differentiating displacement with respect to time: \(v = \frac{ds}{dt} = 16t + 3\). For initial velocity, substitute \(t = 0\), which gives \(v = 3\text{ cm/s}\).

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Question 45: easy

A particle moves in x-y plane according to rule \(x = a \sin \omega t\) and \(y = a \cos \omega t\). The particle follows :

1. An elliptical path
2. A circular path
3. A parabolic path
4. A straight line path equally inclined to x and y-axes
View Answer

Squaring and adding the coordinates: \(x^2 + y^2 = a^2 \sin^2\omega t + a^2 \cos^2\omega t = a^2\). This is the standard equation of a circle of radius \(a\).

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Question 46: easy

The x and y co-ordinates of a particle at any time \(t\) are given by : \(x = 7t + 4t^2\) and \(y = 5t\) where \(x\) and \(y\) are in m and \(t\) in s. The acceleration of the particle at 5 s is :-

1. Zero
2. \( 8 m/s^{2}\)
3. \(20 m/s^{2}\)
4. \(40 m/s^{2} \)
View Answer

Velocity components: \(v_x = \frac{dx}{dt} = 7 + 8t\) and \(v_y = \frac{dy}{dt} = 5\). Acceleration components: \(a_x = \frac{dv_x}{dt} = 8\) and \(a_y = \frac{dv_y}{dt} = 0\). Total acceleration is \(a = \sqrt{a_x^2 + a_y^2} = 8\text{ m/s}^2\) at any time.

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Question 47: easy

A particle moving with a velocity equal to 0.4 m/s is subjected to an uniform acceleration of

\( 0.15 m/s^{2} \) for 2 sec in a direction at right angles to its initial direction of motion. The resultant velocity is :

1. 0.7 m/s
2. 0.5 m/s
3. 0.1 m/s
4. None of these
View Answer

Initial velocity along x-axis is \(v_x = 0.4\text{ m/s}\). Velocity developed along the perpendicular y-axis is \(v_y = a_y t = 0.15 \times 2 = 0.3\text{ m/s}\). Resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\text{ m/s}\).

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Question 48: easy

The motion of a body falling from rest in a viscous medium is described by \(\frac{dv}{dt} = A – Bv\), where A and B are constants. The velocity at time t is given by :

1. \( \frac{A}{B}(1 - e^{-Bt})\)
2. \( A(1 - e^{-B^{2}t}) \)
3. \( ABe^{-t} \)
4. \( AB^{2}(1 - t) \)
View Answer

Integrating the equation \(\int_0^v \frac{dv}{A - Bv} = \int_0^t dt\) gives \(-\frac{1}{B} \ln(\frac{A - Bv}{A}) = t\). Simplifying for velocity gives \(v = \frac{A}{B}(1 - e^{-Bt})\).

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Question 49: easy

Two bodies of different masses \(m_a\) and \(m_b\) are dropped from two different heights, viz, an 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. \( \sqrt{a} : \sqrt{b} \)
4. \(a^2 : b^2 \)
View Answer

The time taken to fall through height \(h\) from rest under gravity is \(t = \sqrt{\frac{2h}{g}}\). Since \(t \propto \sqrt{h}\), the ratio of times for heights \(a\) and \(b\) is \(\sqrt{a} : \sqrt{b}\).

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Question 50: easy

A particle moves along the parabolic path \(y = ax^2\) in such a way that x-component of velocity remains constant, say c. The acceleration of the particle is :

1. \( 2a^{2}c\hat{j} \)
2. \( 2ac^{2}\hat{j} \)
3. \( ac\hat{j} \)
4. \( a^{2}c^{2}\hat{j} \)
View Answer

Given \(v_x = \frac{dx}{dt} = c\) (constant), hence \(a_x = 0\). Differentiating \(y = ax^2\) gives \(v_y = 2ax\frac{dx}{dt} = 2axc\). Differentiating again, \(a_y = 2ac\frac{dx}{dt} = 2ac^2\). Therefore, total acceleration vector is \(a = 2ac^2\hat{j}\).

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