JEE Physics Kinematics MCQs Set 08

Multiple Choice Questions (MCQs) for JEE Physics: Kinematics

Review structured MCQ sets for JEE Physics Kinematics. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Kinematics MCQs for JEE Physics

Access the complete set of multiple-choice questions for Kinematics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: A wooden block of mass 10 g is dropped from the top of a 100 m high cliff. At the same instant a bullet of mass 10 g is fired upward from the foot of the cliff with speed 100 m/s. After what time do they meet?
(a) 1 s
(b) 2 s
(c) 4 s
(d) 10 s
Show Answer & Explanation

Answer: (a) 1 s

Explanation:
1. Take upward positive. Bullet: \( y_1=100t-\tfrac12gt^2 \); block: \( y_2=100-\tfrac12gt^2 \).
2. They meet when \( y_1=y_2 \): \( 100t=100 \), so \( t=1 \) s.

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Question 2: A stone is thrown horizontally with speed \( \sqrt{2gh} \) from the top of a wall of height h. The distance from the wall at which it reaches the ground is
(a) 2h
(b) h
(c) \( \sqrt2\,h \)
(d) 4h
Show Answer & Explanation

Answer: (a) 2h

Explanation:
1. Time of fall: \( h=\tfrac12gt^2 \) gives \( t=\sqrt{2h/g} \).
2. Horizontal distance \( =\sqrt{2gh}\times\sqrt{2h/g}=2h \).

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Question 3: Two trains A and B, each 400 m long, move on parallel tracks at 72 km/h in the same direction, with A ahead. B accelerates at 1 m/s\(^2\) and after 50 s the guard of B just brushes past the driver of A. The original distance between the guard of B and the driver of A was
(a) 400 m
(b) 850 m
(c) 1250 m
(d) 2500 m
Show Answer & Explanation

Answer: (c) 1250 m

Explanation:
1. In the frame of A, B starts at rest and accelerates at 1 m/s\(^2\).
2. Relative displacement in 50 s \( =\tfrac12(1)(50)^2=1250 \) m.
3. The guard of B, at the rear of B, reaches the driver of A at the front of A after covering this relative distance, so the initial gap is 1250 m.

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Question 4: A car moves at 30 m/s on a circular path of radius 500 m. Its speed increases at 2 m/s\(^2\). The magnitude of its acceleration is
(a) 2 m/s\(^2\)
(b) about 2.7 m/s\(^2\)
(c) 1.8 m/s\(^2\)
(d) 3.8 m/s\(^2\)
Show Answer & Explanation

Answer: (b) about 2.7 m/s\(^2\)

Explanation:
1. Centripetal acceleration \( a_c=v^2/r=900/500=1.8 \) m/s\(^2\).
2. Tangential acceleration \( a_t=2 \) m/s\(^2\).
3. Total \( a=\sqrt{1.8^2+2^2}=\sqrt{7.24}\approx2.7 \) m/s\(^2\).

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Question 5: A particle is thrown vertically upward. Its speed at half of the maximum height is 10 m/s. The maximum height attained is (g = 10 m/s\(^2\))
(a) 5 m
(b) 15 m
(c) 20 m
(d) 10 m
Show Answer & Explanation

Answer: (d) 10 m

Explanation:
1. At height \( H/2 \): \( v^2=u^2-2g(H/2)=u^2-gH \).
2. Since \( u^2=2gH \), \( v^2=gH \).
3. \( 100=10H \), so \( H=10 \) m.

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Question 6: A car moving at 40 km/h can be stopped by applying brakes in 2 m. If the same car moves at 80 km/h, the minimum stopping distance is
(a) 8 m
(b) 4 m
(c) 16 m
(d) 6 m
Show Answer & Explanation

Answer: (a) 8 m

Explanation:
1. With constant retardation, \( s=u^2/2a \), so \( s\propto u^2 \).
2. Doubling the speed makes the distance four times: \( 4\times2=8 \) m.

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Question 7: The position of a particle on the x-axis is \( x=a+bt^2 \), with a = 4.5 m and b = 3.5 m/s\(^2\). Its average velocity between t = 1 s and t = 3 s is
(a) 7.0 m/s
(b) 14.0 m/s
(c) 21.0 m/s
(d) 10.5 m/s
Show Answer & Explanation

Answer: (b) 14.0 m/s

Explanation:
1. \( x(1)=4.5+3.5=8 \) m and \( x(3)=4.5+31.5=36 \) m.
2. Average velocity \( =(36-8)/2=14 \) m/s.

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Question 8: A body dropped from height h with zero initial velocity strikes the ground at 3 m/s. Another body is dropped from the same height with an initial velocity of 4 m/s. It strikes the ground with a velocity of
(a) 7 m/s
(b) 1 m/s
(c) 3.5 m/s
(d) 5 m/s
Show Answer & Explanation

Answer: (d) 5 m/s

Explanation:
1. \( v^2=2gh=9 \), so \( 2gh=9 \).
2. For the second body \( v^2=4^2+2gh=16+9=25 \), so \( v=5 \) m/s.

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Question 9: A car accelerates from rest at a constant rate \( \alpha \) for some time, then decelerates at a constant rate \( \beta \) and comes to rest. If the total time is t, the maximum velocity acquired is
(a) \( \dfrac{\alpha\beta\,t}{\alpha+\beta} \)
(b) \( \dfrac{(\alpha+\beta)t}{\alpha\beta} \)
(c) \( \dfrac{(\alpha+\beta)t}{2} \)
(d) \( \dfrac{\alpha t}{\alpha+\beta} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{\alpha\beta\,t}{\alpha+\beta} \)

Explanation:
1. Let the maximum velocity be v. Acceleration time \( t_1=v/\alpha \); deceleration time \( t_2=v/\beta \).
2. \( t_1+t_2=t \), so \( v\left(\dfrac1\alpha+\dfrac1\beta\right)=t \).
3. Hence \( v=\dfrac{\alpha\beta t}{\alpha+\beta} \).

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Question 10: An aeroplane flying horizontally at 216 km/h at a height of 1960 m releases a bomb when vertically above a point A on the ground. The bomb strikes the ground at B. The distance AB is (g = 9.8 m/s\(^2\))
(a) 600 m
(b) 2400 m
(c) 1200 m
(d) 1960 m
Show Answer & Explanation

Answer: (c) 1200 m

Explanation:
1. Speed \( =216\times\tfrac{5}{18}=60 \) m/s. Fall time: \( 1960=\tfrac12(9.8)t^2 \) gives \( t=20 \) s.
2. \( AB=60\times20=1200 \) m.

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Question 11: A body is projected horizontally at 20 m/s from the top of a tower. Its speed after 5 s is nearly (g = 10 m/s\(^2\))
(a) 54 m/s
(b) 50 m/s
(c) 70 m/s
(d) 45 m/s
Show Answer & Explanation

Answer: (a) 54 m/s

Explanation:
1. \( v_x=20 \) m/s and \( v_y=gt=50 \) m/s.
2. \( v=\sqrt{20^2+50^2}=\sqrt{2900}\approx54 \) m/s.

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Question 12: A bus covers 200 m. It covers the first half of the distance at 40 km/h and the second half at speed v. The average speed is 48 km/h. The value of v is
(a) 60 km/h
(b) 56 km/h
(c) 48 km/h
(d) 72 km/h
Show Answer & Explanation

Answer: (a) 60 km/h

Explanation:
1. Average speed for equal distances is the harmonic mean: \( \dfrac{2\cdot40\,v}{40+v}=48 \).
2. \( 80v=1920+48v \), so \( v=60 \) km/h.

Teacher's Note:
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138. .

Question 13: The position of a particle along the x-axis is \( x=9.75+1.50t^3 \) cm (t in s). The average velocity between t = 2 s and t = 3 s is
(a) 28.5 cm/s
(b) 18.0 cm/s
(c) 30.4 cm/s
(d) 22.5 cm/s
Show Answer & Explanation

Answer: (a) 28.5 cm/s

Explanation:
1. \( x(3)-x(2)=1.5(27-8)=28.5 \) cm in 1 s.
2. Average velocity \( =28.5 \) cm/s.

Teacher's Note:
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Question 14: For the particle \( x=9.75+1.50t^3 \) cm, the instantaneous velocity at t = 2 s is
(a) 18.0 cm/s
(b) 28.5 cm/s
(c) 9.0 cm/s
(d) 13.5 cm/s
Show Answer & Explanation

Answer: (a) 18.0 cm/s

Explanation:
1. \( v=dx/dt=4.5t^2 \).
2. At \( t=2 \) s, \( v=4.5\times4=18 \) cm/s.

Teacher's Note:
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Question 15: A train starts from rest with constant acceleration. At one instant its speed is 33.0 m/s and 160 m farther it is 54.0 m/s. The acceleration is
(a) 5.71 m/s\(^2\)
(b) 2.85 m/s\(^2\)
(c) 6.5 m/s\(^2\)
(d) 4.2 m/s\(^2\)
Show Answer & Explanation

Answer: (a) 5.71 m/s\(^2\)

Explanation:
1. \( v^2-u^2=2as \): \( 54^2-33^2=2a(160) \).
2. \( 2916-1089=1827=320a \), so \( a=5.71 \) m/s\(^2\).

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Question 16: In the situation above (the train with a = 5.71 m/s\(^2\) starting from rest), the distance moved from rest until the train reached 33.0 m/s is
(a) 95.4 m
(b) 47.7 m
(c) 190.8 m
(d) 160 m
Show Answer & Explanation

Answer: (a) 95.4 m

Explanation:
1. \( s=v^2/2a=1089/(2\times5.71) \).
2. \( s\approx95.4 \) m.

Teacher's Note:
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Question 17: A body moving in a straight line with constant retardation travels 2 m in the first two seconds and 2.2 m in the next four seconds. Its velocity at the end of the seventh second from the start is
(a) 0.1 m/s
(b) 0.5 m/s
(c) 0 m/s
(d) 1.15 m/s
Show Answer & Explanation

Answer: (a) 0.1 m/s

Explanation:
1. Use \( s=ut-\tfrac12at^2 \). \( 2u-2a=2 \) and \( 6u-18a=4.2 \).
2. From the first, \( u=1+a \); then \( 6+6a-18a=4.2 \), so \( a=0.15 \) m/s\(^2\) and \( u=1.15 \) m/s.
3. \( v(7)=1.15-0.15\times7=0.1 \) m/s.

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Question 18: A motorcyclist moving with uniform retardation takes 10 s and 20 s to cover successive quarter-kilometres. How much farther does he travel after the second quarter-kilometre before coming to rest?
(a) about 10.4 m
(b) about 25 m
(c) about 5 m
(d) about 20.8 m
Show Answer & Explanation

Answer: (a) about 10.4 m

Explanation:
1. Distance 250 m in 10 s and 500 m in 30 s: \( 250=10u-50a \) and \( 500=30u-450a \).
2. Solving, \( a=\tfrac{5}{6} \) m/s\(^2\) and \( u=\tfrac{175}{6}\approx29.17 \) m/s.
3. Total stopping distance \( =u^2/2a\approx510.4 \) m, so the extra distance is about 10.4 m.

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Question 19: A car moves at 20 m/s on a straight road. At t = 0 the driver sees an obstacle 150 m ahead and applies brakes giving a retardation of 2 m/s\(^2\). The distance of the car from the obstacle at t = 15 s is
(a) 50 m
(b) 150 m
(c) 0 m (it hits the obstacle)
(d) 25 m
Show Answer & Explanation

Answer: (a) 50 m

Explanation:
1. Time to stop: \( t=20/2=10 \) s, so the car is already at rest at 15 s.
2. Stopping distance \( =20^2/(2\times2)=100 \) m.
3. Distance from the obstacle \( =150-100=50 \) m.

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Question 20: A ball is thrown at 100 m/s at 30\(^\circ\) to the horizontal from a plane and meets the same plane again. Take g = 9.8 m/s\(^2\). Its time of flight and horizontal range are
(a) 5.10 s and 441.8 m
(b) 20.4 s and 1767 m
(c) 10.20 s and 883.7 m
(d) 10.20 s and 1020 m
Show Answer & Explanation

Answer: (c) 10.20 s and 883.7 m

Explanation:
1. \( T=2u\sin\theta/g=2(100)(0.5)/9.8=10.20 \) s.
2. \( R=u^2\sin2\theta/g=10^4(0.866)/9.8=883.7 \) m.

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Practice MCQs for JEE Physics Kinematics

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FAQs

Where can I access latest JEE Physics Kinematics MCQs Set 08?

You can get most exhaustive JEE Physics Kinematics MCQs Set 08 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Kinematics MCQs Set 08 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Kinematics MCQs Set 08, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Kinematics MCQs Set 08?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Kinematics MCQs Set 08 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.