NEET Equations of Motion MCQs & PYQs

Practice NEET Equations of Motion Questions

Question 1:

moderate

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

(1994)

Concept: Kinematics equations involving differentiation.
Formula: Velocity \(v = \frac{ds}{dt}\) and acceleration \(a = \frac{dv}{dt}\).
Solution: Given \(s = t^3 - 6t^2 + 3t + 4\). Then \(v = 3t^2 - 12t + 3\) and \(a = 6t - 12\). Setting \(a=0\) gives \(t=2\text{ s}\). Substituting \(t=2\text{ s}\) into \(v\) gives \(v = 3(2)^2 - 12(2) + 3 = 12 - 24 + 3 = -9\text{ m/s}\).

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Question 2:

easy

A body starts from rest, what is the ratio of the distance travelled by the body during the \(4^{\text{th}}\) and \(3^{\text{rd}}\) second?

(1993)

Concept: Distance covered in the \(n^{text{th}}\) second for uniformly accelerated motion.
Formula: \(S_n = u + \frac{a}{2}(2n - 1)\). Since it starts from rest, \(u=0\).
Solution: \(S_4 = \frac{a}{2}(2(4) - 1) = \frac{7a}{2}\), \(S_3 = \frac{a}{2}(2(3) - 1) = \frac{5a}{2}\). Ratio \(S_4:S_3 = 7a/2 : 5a/2 = 7:5\).

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Question 3:

moderate

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity \(30\text{ km/h}\) and \(40\text{ km/h}\) respectively. The velocity of the car midway between P and Q is:

(1988)

Concept: Equations of motion under uniform acceleration.
Formula: \(v^2 = u^2 + 2as\).
Solution: Let \(u_P=30\), \(v_Q=40\) and distance be \(s\). \(v_Q^2 = u_P^2 + 2as\) gives \(40^2 = 30^2 + 2as\) => \(1600 = 900 + 2as\) => \(2as = 700\) => \(as = 350\). For the midway point, \(v_m^2 = u_P^2 + 2a(s/2) = u_P^2 + as = 30^2 + 350 = 900 + 350 = 1250\). So, \(v_m = sqrt{1250} = 25\sqrt{2}\text{ km/h}\).

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Question 4:

moderate

The ratio of the distance traveled by a freely falling body in the \(1^{text{st}}\,\text{ }2^{text{nd}}\,\text{ }3^{text{rd}}\) and \(4^{text{th}}\) second:

(2022)

Concept: Galileo's law of odd numbers for free fall.
Formula: Distance in \(n^{text{th}}\) second is \(S_n = \frac{g}{2}(2n - 1)\).
Solution: \(S_1:S_2:S_3:S_4 = (2(1)-1):(2(2)-1):(2(3)-1):(2(4)-1) = 1:3:5:7\).

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Question 5:

difficult

A ball is dropped from a high rise platform at \(t = 0\) starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed \(v\). The two balls meet at \(t = 18\text{s}\). What is the value of \(v\)?

[2010 Pre]

Concept: Motion under gravity and meeting condition.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: For ball 1 (dropped at \(t=0\)): \(h_1 = \frac{1}{2}g(18)^2 = 162g\). For ball 2 (thrown at \(t=6\text{ s}\), travels for \(12\text{ s}\)): \(h_2 = v(12) + \frac{1}{2}g(12)^2 = 12v + 72g\). When they meet, \(h_1 = h_2\): \(162g = 12v + 72g\). \(90g = 12v\). Using \(g=10\text{ m/s}^2\), \(900 = 12v\) => \(v = 75\text{ m/s}\).

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Question 6:

moderate

A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10\) seconds is \(S_1\) and that covered in the first \(20\)seconds is \(S_2\), then:

(2009)

For constant acceleration from rest, distance \(S = \frac{1}{2}at^2\). So \(S \propto t^2\). For \(t=10 \text{ s}\), \(S_1 = \frac{1}{2}a(10)^2 = 50a\). For \(t=20 \text{ s}\), \(S_2 = \frac{1}{2}a(20)^2 = 200a\). Therefore, \(S_2 = 4S_1\).

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Question 7:

moderate

The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \text{ m s}^{-2}\), in the third second is

(2008)

The distance in the \(n^{\text{th}}\) second is given by \(S_n = u + \frac{a}{2}(2n - 1)\). Here (u=0), \(a = \frac{4}{3} \text{ m s}^{-2}\) and (n=3). So \(S_3 = 0 + \frac{4/3}{2}(2 times 3 - 1) = \frac{2}{3}(5) = \frac{10}{3} \text{ m}\).

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Question 8:

moderate

A particle moves in a straight line with a constant acceleration. It changes its velocity from \(10 \text{ m s}^{-1}\) to \(20 \text{ m s}^{-1}\) while passing through a distance \(135 \text{ m}\) in (t) second. The value of (t) is

(2008)

Given \(u = 10 \text{ m/s}\), \(v = 20 \text{ m/s}\), (\S = 135 \text{ m}\). Using \(v^2 = u^2 + 2aS\), \((20)^2 = (10)^2 + 2a(135) \Rightarrow 400 = 100 + 270a \Rightarrow a = \frac{300}{270} = \frac{10}{9} \text{ m/s}^2\). Now use \(v = u + at\), \(20 = 10 + \frac{10}{9}t \Rightarrow 10 = \frac{10}{9}t \Rightarrow t = 9 \text{ s}\).

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Question 9:

easy

If a car at rest accelerates uniformly to a speed of \(144 \text{ km/h}\) in \(20 \text{ sec}\), it covers a distance of:

(1997)

Given (u=0), \(v = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}\), \(t = 20 \text{ s}\). Using \(S = \frac{u+v}{2}t), we get \(S = \frac{0+40}{2} \times 20 = 20 \times 20 = 400 \text{ m}\).

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Question 10:

easy

The position (x) of a particle varies with time, (t), as \(x = at^2 – bt^3\). The acceleration will be zero at time (t) equal to:

(1997)

Given \(x = at^2 - bt^3\). Velocity \(v = \frac{dx}{dt} = 2at - 3bt^2\). Acceleration \(a_c = \frac{dv}{dt} = 2a - 6bt\). For zero acceleration, \(2a - 6bt = 0 \Rightarrow 2a = 6bt \Rightarrow t = \frac{2a}{6b} = \frac{a}{3b}\).

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