50+ JEE Mains MCQ Questions Motion in a Straight Line with Solutions

Preparing for the Joint Entrance Exam (JEE) can be a daunting task. With so many subjects to cover and so many topics to study, it can be challenging to know where to start. One essential topic in the JEE Mains syllabus is the Motion in a Straight Line. In this article, we will provide 50+ MCQ questions on the Motion in a Straight Line, along with detailed solutions to help you prepare for the JEE Mains exam.

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These 50+ MCQ questions are selected by the experts of studyrate.in and these are more difficult questions, which will help you to better understand Motion in a Straight Line JEE Mains MCQ Questions with Answers.

Motion in a Straight Line JEE Mains MCQ

Which of the following is a scalar quantity?
a) Velocity
b) Displacement
c) Acceleration
d) Speed

Answer: d) Speed


A particle starts from rest and moves along a straight line with an acceleration given by a = (4t^2 – 3t + 2) m/s^2. What is the total displacement of the particle in the first 2 seconds?
a) 2 m
b) 4 m
c) 6 m
d) 8 m

Answer: c) 6 m Explanation: To find the displacement, we need to integrate the given acceleration function with respect to time. Let’s solve it step by step: Given acceleration, a = 4t^2 – 3t + 2 Integrating acceleration with respect to time, we get: v = ∫(4t^2 – 3t + 2) dt Integrating the terms separately: v = (4/3)t^3 – (3/2)t^2 + 2t + C Since the particle starts from rest, the constant of integration (C) is zero. So, v = (4/3)t^3 – (3/2)t^2 + 2t Now, to find the displacement, we need to integrate the velocity function: s = ∫[(4/3)t^3 – (3/2)t^2 + 2t] dt Integrating the terms separately: s = (1/3)t^4 – (1/2)t^3 + t^2 + D Since we are interested in finding the displacement in the first 2 seconds, we evaluate the displacement function at t = 2: s = (1/3)(2)^4 – (1/2)(2)^3 + (2)^2 + D s = (16/3) – (8/2) + 4 + D s = (16/3) – 8 + 4 + D s = 16/3 – 4 + D s = 16/3 – 12/3 + D s = 4/3 + D Since the constant of integration (D) can take any value, we cannot determine the exact displacement. However, we can determine the change in displacement between two time instants. In this case, the change in displacement between t = 0 and t = 2 is: Δs = s(t = 2) – s(t = 0) Δs = (4/3 + D) – (4/3 + D) Δs = 0

A particle starts from rest and moves along a straight line with a velocity given by v = 10t – 5t^2, where v is in m/s and t is in seconds. What is the distance traveled by the particle when its velocity becomes zero?
a) 10 m
b) 12 m
c) 14 m
d) 16 m

Answer: b) 12 m Explanation: To find the distance traveled by the particle, we need to find the total area under the velocity-time curve when the velocity becomes zero. Let’s solve it step by step: Given velocity, v = 10t – 5t^2 To find the time at which the velocity becomes zero, we set v = 0 and solve for t: 10t – 5t^2 = 0 Factorizing the equation, we get: t(10 – 5t) = 0 This equation gives us two solutions: t = 0 (at the starting point) and t = 2 seconds So, the particle’s velocity becomes zero at t = 2 seconds. To find the distance traveled by the particle, we need to calculate the area under the velocity-time curve between t = 0 and t = 2 seconds: Distance = ∫(10t – 5t^2) dt (from t = 0 to t = 2) Integrating the terms separately: Distance = ∫(10t) dt – ∫(5t^2) dt (from t = 0 to t = 2) Distance = 5t^2 – (5/3)t^3 (from t = 0 to t = 2) Calculating the distance: Distance = 5(2)^2 – (5/3)(2)^3 – (5(0)^2 – (5/3)(0)^3) Distance = 20 – (40/3) – 0 Distance = 20 – (40/3) Distance = (60/3) – (40/3) Distance = 20/3 Therefore, the distance traveled by the particle when its velocity becomes zero is 20/3 meters, which is approximately equal to 6.67 meters.


A car moves along a straight road such that its acceleration at any time t is given by a = 4t^3 – 6t^2 + 2t. If the car starts from rest, what is the distance covered by the car in the time interval from t = 0 to t = 2 seconds?
a) 4 m
b) 8 m
c) 12 m
d) 16 m

Answer: c) 12 m Explanation: To find the distance covered by the car, we need to integrate the velocity function with respect to time. Let’s solve it step by step: Given acceleration, a = 4t^3 – 6t^2 + 2t Integrating acceleration with respect to time, we get: v = ∫(4t^3 – 6t^2 + 2t) dt Integrating the terms separately: v = (1/4)t^4 – (2/3)t^3 + t^2 + C Since the car starts from rest, the constant of integration (C) is zero. So, v = (1/4)t^4 – (2/3)t^3 + t^2 Now, to find the distance, we need to integrate the velocity function: s = ∫[(1/4)t^4 – (2/3)t^3 + t^2] dt Integrating the terms separately: s = (1/20)t^5 – (1/12)t^4 + (1/3)t^3 + D Since we are interested in finding the distance covered by the car in the time interval from t = 0 to t = 2 seconds, we evaluate the displacement function at t = 2 and t = 0: s = [(1/20)(2)^5 – (1/12)(2)^4 + (1/3)(2)^3 + D] – [(1/20)(0)^5 – (1/12)(0)^4 + (1/3)(0)^3 + D] s = [(1/20)(32) – (1/12)(16) + (1/3)(8) + D] – [0 – 0 + 0 + D] s = [32/20 – 16/12 + 8/3 + D] – [0 + 0 + 0 + D] s = [8/5 – 4/3 + 8/3 + D] – [0 + 0 + 0 + D] s = (24 – 20 + 40)/15 s = 44/15 Therefore, the distance covered by the car in the time interval from t = 0 to t = 2 seconds is 44/15 meters, which is approximately equal to 2.93 meters.


A particle moves along a straight line such that its displacement at any time t is given by s = 3t^3 – 2t^2 + 5t + 1. What is the acceleration of the particle?
a) 9t^2 – 4t + 5
b) 6t^2 – 4t + 5
c) 3t^2 – 2t + 5
d) 6t – 2

Answer: a) 9t^2 – 4t + 5

A car travels along a straight road with a velocity of 40 m/s. The car decelerates uniformly at a rate of 8 m/s^2 until it comes to rest. How far does the car travel during this deceleration?
a) 160 m
b) 200 m
c) 240 m
d) 320 m

Answer: b) 200 m

A particle moves along a straight line according to the equation x = 5t^2 – 3t + 2, where x is the position of the particle and t is the time. What is the average velocity of the particle over the time interval from t = 1 to t = 3 seconds?
a) 20 m/s
b) 24 m/s
c) 28 m/s
d) 32 m/s

Answer: c) 28 m/s

A car travels along a straight road with a velocity given by v = 4t^2 – 6t + 3, where v is in m/s and t is in seconds. What is the distance covered by the car in the time interval from t = 0 to t = 2 seconds?
a) 2 m
b) 4 m
c) 6 m
d) 8 m

Answer: b) 4 m

A particle is moving along a straight line with an initial velocity of 10 m/s. The particle undergoes a constant acceleration of 2 m/s^2. How long will it take for the particle to come to rest?
a) 5 s
b) 10 s
c) 15 s
d) 20 s

Answer: a) 5 s

A car travels along a straight road with a velocity given by v = 8t – 3t^2, where v is in m/s and t is in seconds. At what time does the car have maximum velocity?
a) 0.5 s
b) 1 s
c) 1.5 s
d) 2 s

Answer: a) 5 s

A particle moves along a straight line with an initial velocity of 10 m/s. Its acceleration is given by a = 2t – 1, where a is in m/s^2 and t is in seconds. What is the displacement of the particle at t = 4 seconds?
a) 48 m
b) 56 m
c) 64 m
d) 72 m

Answer: c) 64 m

A ball is thrown vertically upwards with an initial velocity of 20 m/s. What is the maximum height reached by the ball? (Assume g = 10 m/s^2)
a) 20 m
b) 40 m
c) 60 m
d) 80 m

Answer: c) 60 m

A particle moves along a straight line such that its displacement at any time t is given by s = 2t^3 – 3t^2 + 6t – 4. What is the average speed of the particle in the time interval from t = 1 to t = 2 seconds?
a) 16 m/s
b) 14 m/s
c) 12 m/s
d) 10 m/s

Answer: b) 14 m/s

A car is traveling along a straight road with a constant speed of 60 km/h. It comes to a stop in 10 seconds. What is the acceleration of the car?
a) 2 m/s^2
b) 3 m/s^2
c) 4 m/s^2
d) 5 m/s^2

Answer: a) 2 m/s^2

A particle moves along a straight line according to the equation x = 4t^2 – 6t + 3, where x is the position of the particle and t is the time. What is the instantaneous velocity of the particle at t = 2 seconds?
a) 10 m/s
b) 12 m/s
c) 14 m/s
d) 16 m/s

Answer: b) 12 m/s

A car travels along a straight road with a velocity given by v = 10t – 4t^2, where v is in m/s and t is in seconds. At what time does the car have maximum acceleration?
a) 0.5 s
b) 1 s
c) 1.5 s
d) 2 s

Answer: c) 1.5 s

A particle moves along a straight line with an initial velocity of 5 m/s. Its acceleration is given by a = 4t + 2, where a is in m/s^2 and t is in seconds. What is the velocity of the particle at t = 3 seconds?
a) 23 m/s
b) 25 m/s
c) 27 m/s
d) 29 m/s

Answer: c) 27 m/s

A ball is thrown vertically upwards with an initial velocity of 30 m/s. What is the time taken for the ball to reach its maximum height? (Assume g = 10 m/s^2)
a) 1 s
b) 2 s
c) 3 s
d) 4 s

Answer: b) 2 s

A particle moves along a straight line such that its displacement at any time t is given by s = 5t^2 – 4t + 3. What is the velocity of the particle at t = 2 seconds?
a) 14 m/s
b) 16 m/s
c) 18 m/s
d) 20 m/s

Answer: b) 16 m/s

A car is traveling along a straight road with a velocity given by v = 20t – 3t^2, where v is in m/s and t is in seconds. At what time does the car have zero acceleration?
a) 1 s
b) 2 s
c) 3 s d) 4 s

Answer: c) 3 s

A particle moves along a straight line with an initial velocity of 8 m/s. Its acceleration is given by a = 3t^2 – 2, where a is in m/s^2 and t is in seconds. What is the displacement of the particle at t = 5 seconds?
a) 176 m
b) 184 m
c) 192 m
d) 200 m

Answer: c) 3 s

A ball is thrown vertically upwards with an initial velocity of 15 m/s. What is the total time of flight of the ball? (Assume g = 10 m/s^2)
a) 3 s
b) 4 s
c) 5 s
d) 6 s

Answer: c) 5 s

A particle moves along a straight line with a velocity given by v = 3t^2 – 2t + 5, where v is in m/s and t is in seconds. What is the acceleration of the particle when t = 2 seconds?
a) 10 m/s^2
b) 12 m/s^2
c) 14 m/s^2
d) 16 m/s^2

Answer: b) 12 m/s^2

A car is traveling along a straight road with a constant acceleration of 4 m/s^2. If its initial velocity is 20 m/s, what is its velocity after 5 seconds?
a) 40 m/s
b) 60 m/s
c) 80 m/s
d) 100 m/s

Answer: c) 80 m/s

A particle moves along a straight line with a velocity given by v = 6t – 2t^2, where v is in m/s and t is in seconds. What is the displacement of the particle when its acceleration becomes zero?
a) 2 m
b) 4 m
c) 6 m
d) 8 m

Answer: b) 4 m

A car is traveling along a straight road with a velocity given by v = 12t – 3t^2, where v is in m/s and t is in seconds. What is the distance covered by the car in the time interval from t = 0 to t = 3 seconds?
a) 18 m
b) 24 m
c) 30 m
d) 36 m

Answer: d) 36 m

A particle is moving along a straight line with an initial velocity of 10 m/s. Its acceleration is given by a = 2t + 3, where a is in m/s^2 and t is in seconds. What is the velocity of the particle at t = 4 seconds?
a) 23 m/s
b) 25 m/s
c) 27 m/s
d) 29 m/s

Answer: d) 29 m/s

A ball is thrown vertically upwards with an initial velocity of 25 m/s. What is the maximum height reached by the ball? (Assume g = 10 m/s^2)
a) 20 m
b) 30 m
c) 40 m
d) 50 m

Answer: d) 50 m

A particle moves along a straight line such that its displacement at any time t is given by s = 3t^3 – 2t^2 + 4t + 1. What is the average speed of the particle in the time interval from t = 1 to t = 3 seconds?
a) 14 m/s
b) 16 m/s
c) 18 m/s
d) 20 m/s

Answer: b) 16 m/s

A car is traveling along a straight road with a constant speed of 72 km/h. It comes to a stop in 15 seconds. What is the acceleration of the car?
a) 1 m/s^2
b) 2 m/s^2
c) 3 m/s^2
d) 4 m/s^2

Answer: c) 3 m/s^2

A particle moves along a straight line according to the equation x = 3t^2 – 2t + 5, where x is the position of the particle and t is the time. What is the instantaneous velocity of the particle at t = 3 seconds?
a) 22 m/s
b) 24 m/s
c) 26 m/s
d) 28 m/s

Answer: c) 26 m/s

A car travels along a straight road with a velocity given by v = 5t^2 – 4t, where v is in m/s and t is in seconds. At what time does the car have maximum acceleration?
a) 0.5 s
b) 1 s
c) 1.5 s
d) 2 s

Answer: c) 1.5 s

A particle moves along a straight line with an initial velocity of 6 m/s. Its acceleration is given by a = 3t^2 – 2, where a is in m/s^2 and t is in seconds. What is the displacement of the particle at t = 5 seconds?
a) 176 m
b) 184 m
c) 192 m
d) 200 m

Answer: b) 184 mA ball is thrown vertically upwards with an initial velocity of 20 m/s.

What is the time taken for the ball to reach half of its maximum height? (Assume g = 10 m/s^2)
a) 1 s
b) 2 s
c) 3 s
d) 4 s

Answer: b) 2 s

We hope there JEE MCQ of Class 11 Motion in a Straight Line will help you to score an excellent rank in JEE Mains and Advanced. If you have any queries feel free to write in the comments section. We at Study Rate are always ready to serve our students.

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