Equations of Motion - NEET Physics Chapterwise MCQs & PYQs

NEET Equations of Motion MCQs & PYQs

Practice NEET Equations of Motion Questions

Question 11:

difficult

The acceleration of a particle is increasing linearly with time (t) as \(bt\). The particle starts from origin with an initial velocity \(v_0\). The distance travelled by the particle in time (t) will be:

(1995)

Given \(a = \frac{dv}{dt} = bt\). Integrating, \(v = \int bt , dt = \frac{1}{2}bt^2 + C_1\). Since \(v=v_0\) at \(t=0\), \(C_1 = v_0\). So \(v = v_0 + \frac{1}{2}bt^2\). Given \(v = \frac{dx}{dt}\). Integrating again, \(x = \int (v_0 + \frac{1}{2}bt^2\) , \(dt = v_0 t + \frac{1}{2}b \frac{t^3}{3} + C_2\). Since (x=0) at (t=0), (C_2 = 0). Thus, \(x = v_0 t + \frac{bt^3}{6}\).

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Question 12:

moderate

The velocity of train increases uniformly from \(20 \text{ km/h}\) to \(60 \text{ km/h}\) in 4 hours. The distance travelled by the train during this period, is:

(1994)

Given initial velocity \(u = 20 \text{ km/h}\), final velocity \(v = 60 \text{ km/h}\), and time \(t = 4 \text{ h}\). For uniform acceleration, the distance \(S = \frac{u+v}{2}t\). Plugging in the values, \(S = \frac{20 + 60}{2} \times 4 = \frac{80}{2} \times 4 = 40 \times 4 = 160 \text{ km}\).

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Question 13:

easy

What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey?

(1989)

Concept: Distance covered in the \(n^{\text{th}}\)) second of free fall from rest.
Formula: \(h_n = u + \frac{g}{2}(2n-1)\). Since \(u=0\), \(h_n = \frac{g}{2}(2n-1)\).
Distance in 4th second (\(n=4\)): \(h_4 = \frac{g}{2}(2 times 4 - 1) = \frac{7g}{2}\).
Distance in 5th second (\(n=5\)): \(h_5 = \frac{g}{2}(2 times 5 - 1) = \frac{9g}{2}\).
Ratio: \(h_4 : h_5 = \frac{7g}{2} : \frac{9g}{2} = 7:9\).

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Question 14:

difficult

A particle has initial velocity \(3\hat{i} + 4\hat{j}\) and has acceleration \(0.4\hat{i} + 0.3\hat{j}\). Its speed after 10 s is:

(2010 Pre)

Initial velocity \(vec{v}_0 = 3\hat{i} + 4\hat{j}\). Acceleration \(\vec{a} = 0.4\hat{i} + 0.3\hat{j}\). Time \(t = 10\text{ s}\). Using \(\vec{v} = \vec{v}_0 + \vec{a}t\), we get \(\vec{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j})(10) = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) = 7\hat{i} + 7\hat{j}\). Speed is the magnitude of velocity: \(|\vec{v}| = \sqrt{7^2 + 7^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}\).

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Question 15:

moderate

A particle starting from the origin \((0, 0)\) moves in a straight line in the \((x, y)\) plane. Its coordinates at a later time are \((\sqrt{3}, 3)\). The path of the particle makes with the x-axis an angle of:

(2007)

The particle moves from \((0,0)\) to \((\sqrt{3}, 3)\). The slope of this straight line path is \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{\sqrt{3} - 0} = \frac{3}{\sqrt{3}} = \sqrt{3}\). The angle \(theta\) with the x-axis is given by \(tan\theta = m\). So, \(tan\theta = \sqrt{3}\) which implies \(\theta = 60^\circ\).

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Question 16:

difficult

Two boys are standing at the ends A and B of a ground, where \(AB = a\). The boy at B starts running in a direction perpendicular to AB with velocity \(v_1\). The boy at A starts running simultaneously with velocity \(v\) and catches the other boy in a time t, where t is:

(2005)

Let B be at \((0,0)\) and A at \((a,0)\) at \(t=0\). Boy B's position at time \(t\) is \(\vec{r}_B = v_1 t \hat{j}\). Boy A moves with velocity \(\vec{v}_A = v_{Ax}\hat{i} + v_{Ay}\hat{j}\). For A to catch B, their positions must be equal at time \(t\). So, \(a\hat{i} + \vec{v}_A t = v_1 t \hat{j}\). This implies \(v_{Ax} = -a/t\) and \(v_{Ay} = v_1\). The magnitude of A's velocity is \(v = |\vec{v}_A| = \sqrt{v_{Ax}^2 + v_{Ay}^2}\). So, \(v^2 = (-a/t)^2 + v_1^2\). Rearranging for \(t\): \(t^2 = \frac{a^2}{v^2 - v_1^2}\), hence \(t = \frac{a}{\sqrt{v^2 - v_1^2}}\).

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Question 17:

difficult

The position vector of a particle \(\vec{R}\) as a function of time is given by: \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\) Where R is in metres, t is in seconds and \(\hat{i}\) and \(\hat{j}\) denote unit vectors along x and y-direction, respectively. Which one of the following statements is wrong for the motion of particle?

(2015)

From \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\), \(x=4sin(2\pi t)\) and \(y=4cos(2\pi t)\). \(x^2+y^2=16\) implies a circle of radius 4m. \(\vec{V} = 8\pi cos(2\pi t)\hat{i} - 8\pi sin(2\pi t)\hat{j}\). \(|\vec{V}| = 8\pi\text{ m/s}\). \(\vec{a} = -16\pi^2sin(2\pi t)\hat{i} - 16\pi^2cos(2\pi t)\hat{j} = -4\pi^2 \vec{R}\). So \(\vec{a}\) is along \(-\vec{R}\). Also, \(|\vec{a}| = 16\pi^2\) and \(\frac{V^2}{R} = \frac{(8\pi)^2}{4} = 16\pi^2\). Therefore, (a), (b), (c) are correct. (d) is wrong because \(|\vec{V}| = 8\pi\text{ m/s}\), not 8 m/s.

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Question 18:

moderate

A particle has initial velocity \(2\hat{i} + 3\hat{j}\) and acceleration \(0.3\hat{i} + 0.2\hat{j}\) . The magnitude of velocity after 10 sec will be:

(2012 Pre)

Given \(\vec{v}_0 = 2\hat{i} + 3\hat{j}\), \(\vec{a} = 0.3\hat{i} + 0.2\hat{j}\), and \(t = 10\text{ s}\). Using \(\vec{v} = \vec{v}_0 + \vec{a}t\), we get \(\vec{v} = (2\hat{i} + 3\hat{j}) + (0.3\hat{i} + 0.2\hat{j})(10) = (2\hat{i} + 3\hat{j}) + (3\hat{i} + 2\hat{j}) = 5\hat{i} + 5\hat{j}\). The magnitude of velocity is \(|\vec{v}| = \sqrt{5^2 + 5^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}\).

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