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Page 1 JEE Main Previous Year Questions (2021-2025) Motion 2025 Page 2 JEE Main Previous Year Questions (2021-2025) Motion 2025 Q1: A particle initially at rest starts moving from reference point ?? = ?? along ?? -axis, with velocity ?? that varies as ?? = ?? v?? ?? /?? . The acceleration of the particle is ???? -?? . [JEE Main 2024 (Online) 1st February Evening Shift] Ans: 8 To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is ?? = 4v?? However, this function gives the velocity as a function of position ?? , not as a function of time ?? . Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate ?? with respect to ?? . The chain rule in this context can be stated as follows: ?? = ???? ???? = ???? ???? · ???? ???? Now, because ???? ?? ?? is the velocity ?? itself and ???? ???? is the derivative of the velocity with respect to ?? , we first find ???? ???? : ?? = 4v?? = 4?? 1 2 Differentiating with respect to ?? , we get: ???? ???? = 4· 1 2 ?? - 1 2 = 2?? - 1 2 = 2 v?? Now, because ?? = 4v?? , we can rewrite v?? as ?? 4 . Using this to replace v?? in our expression for ???? ???? , we get: ???? ???? = 2 v?? = 2 ?? 4 = 8 ?? Now, using the chain rule: Page 3 JEE Main Previous Year Questions (2021-2025) Motion 2025 Q1: A particle initially at rest starts moving from reference point ?? = ?? along ?? -axis, with velocity ?? that varies as ?? = ?? v?? ?? /?? . The acceleration of the particle is ???? -?? . [JEE Main 2024 (Online) 1st February Evening Shift] Ans: 8 To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is ?? = 4v?? However, this function gives the velocity as a function of position ?? , not as a function of time ?? . Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate ?? with respect to ?? . The chain rule in this context can be stated as follows: ?? = ???? ???? = ???? ???? · ???? ???? Now, because ???? ?? ?? is the velocity ?? itself and ???? ???? is the derivative of the velocity with respect to ?? , we first find ???? ???? : ?? = 4v?? = 4?? 1 2 Differentiating with respect to ?? , we get: ???? ???? = 4· 1 2 ?? - 1 2 = 2?? - 1 2 = 2 v?? Now, because ?? = 4v?? , we can rewrite v?? as ?? 4 . Using this to replace v?? in our expression for ???? ???? , we get: ???? ???? = 2 v?? = 2 ?? 4 = 8 ?? Now, using the chain rule: ?? = ???? ???? = ???? ???? · ???? ???? = 8 ?? · ?? Simplifying this, the velocity terms cancel out, leaving us with: ?? = 8 ms -2 Thus, the acceleration of the particle is 8 ms -2 . Q2: A particle is moving in one dimension (along ?? axis) under the action of a variable force. It's initial position was ???? ?? right of origin. The variation of its position (?? ) with time (?? ) is given as ?? = -?? ?? ?? + ???? ?? ?? + ???? ?? , where ?? is in ?? and ?? is in ?? .The velocity of the particle when its acceleration becomes zero is ?? /?? . [JEE Main 2024 (Online) 1st February Morning Shift] Ans: 52 Position ( x) : The particle's location on the x-axis at a given time. Velocity (v): The rate of change of position with respect to time (?? = ???? ???? ). Acceleration (a): The rate of change of velocity with respect to time (?? = ???? ???? ) The particle's position is given by: ?? (?? )= -3?? 3 + 18?? 2 + 16?? We need to find the velocity when the acceleration is zero. Find the Velocity (v) and Acceleration (a) Functions: Velocity: ?? (?? )= ???? ???? = -9?? 2 + 36?? + 16 Acceleration: ?? (?? )= ???? ???? = -18?? + 36 Find the Time (t) When Acceleration is Zero: ?? (?? )= 0 -18?? + 36 = 0 ?? = 2 seconds Calculate the Velocity at ?? = 2 seconds: ?? (2)= -9(2) 2 + 36(2)+ 16 ?? (2)= 52 m/s Page 4 JEE Main Previous Year Questions (2021-2025) Motion 2025 Q1: A particle initially at rest starts moving from reference point ?? = ?? along ?? -axis, with velocity ?? that varies as ?? = ?? v?? ?? /?? . The acceleration of the particle is ???? -?? . [JEE Main 2024 (Online) 1st February Evening Shift] Ans: 8 To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is ?? = 4v?? However, this function gives the velocity as a function of position ?? , not as a function of time ?? . Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate ?? with respect to ?? . The chain rule in this context can be stated as follows: ?? = ???? ???? = ???? ???? · ???? ???? Now, because ???? ?? ?? is the velocity ?? itself and ???? ???? is the derivative of the velocity with respect to ?? , we first find ???? ???? : ?? = 4v?? = 4?? 1 2 Differentiating with respect to ?? , we get: ???? ???? = 4· 1 2 ?? - 1 2 = 2?? - 1 2 = 2 v?? Now, because ?? = 4v?? , we can rewrite v?? as ?? 4 . Using this to replace v?? in our expression for ???? ???? , we get: ???? ???? = 2 v?? = 2 ?? 4 = 8 ?? Now, using the chain rule: ?? = ???? ???? = ???? ???? · ???? ???? = 8 ?? · ?? Simplifying this, the velocity terms cancel out, leaving us with: ?? = 8 ms -2 Thus, the acceleration of the particle is 8 ms -2 . Q2: A particle is moving in one dimension (along ?? axis) under the action of a variable force. It's initial position was ???? ?? right of origin. The variation of its position (?? ) with time (?? ) is given as ?? = -?? ?? ?? + ???? ?? ?? + ???? ?? , where ?? is in ?? and ?? is in ?? .The velocity of the particle when its acceleration becomes zero is ?? /?? . [JEE Main 2024 (Online) 1st February Morning Shift] Ans: 52 Position ( x) : The particle's location on the x-axis at a given time. Velocity (v): The rate of change of position with respect to time (?? = ???? ???? ). Acceleration (a): The rate of change of velocity with respect to time (?? = ???? ???? ) The particle's position is given by: ?? (?? )= -3?? 3 + 18?? 2 + 16?? We need to find the velocity when the acceleration is zero. Find the Velocity (v) and Acceleration (a) Functions: Velocity: ?? (?? )= ???? ???? = -9?? 2 + 36?? + 16 Acceleration: ?? (?? )= ???? ???? = -18?? + 36 Find the Time (t) When Acceleration is Zero: ?? (?? )= 0 -18?? + 36 = 0 ?? = 2 seconds Calculate the Velocity at ?? = 2 seconds: ?? (2)= -9(2) 2 + 36(2)+ 16 ?? (2)= 52 m/s Answer: The velocity of the particle when its acceleration becomes zero is 52 m/s. Q3: The displacement and the increase in the velocity of a moving particle in the time interval of ?? to (?? + ?? )?? are ?????? ?? and ???? ?? /?? , respectively. The distance travelled by the particle in (?? + ?? ) ???? is ?? . [JEE Main 2024 (Online) 30th January Morning Shift] Ans: 175 Considering acceleration is constant ?? = ?? + ???? ?? + 50 = ?? + ?? ? ?? = 50 m/s 2 125 = ???? + 1 2 ?? ?? 2 125 = ?? + ?? 2 ? ?? = 100 m/s ? ?? ?? 4?? = ?? + ?? 2 [2n - 1] = 175 m Q4: A ball rolls off the top of a stairway with horizontal velocity ?? . The steps are ?? .?? ?? high and ?? .?? ?? wide. The minimum velocity ?? with which that ball just hits the step 5 of the stairway will be v?? ???? -?? where ?? = [use ?? = ???? ?? /?? ?? ]. [JEE Main 2024 (Online) 29th January Morning Shift] Ans: 2 Q5: A body falling under gravity covers two points ?? and ?? separated by ???? ?? in ?? ?? . The distance of upper point A from the starting point is ?? (use ?? = ???? ???? -?? ). [JEE Main 2024 (Online) 27th January Evening Shift] Ans: 45 Q6: A particle starts from origin at ?? = ?? with a velocity ?? ??ˆ ?? /?? and moves in ?? - ?? plane under action of a force which produces a constant acceleration of (?? ??ˆ + ?? ??ˆ)?? /?? ?? . If the ?? -coordinate of the particle at that instant is ???? ?? , then the speed of the particle at this time is v?? ?? /?? . The value of ?? is [JEE Main 2024 (Online) 27th January Morning Shift] Ans: 673 To solve for the value of ?? , which represents the square of the speed of the particle at the time its ?? - coordinate is 84 m, we need to first determine the time at which the particle reaches this ?? coordinate, and then use this time to calculate its final velocity in both the ?? and ?? directions. The motion of the particle in the ?? -direction can be described by the kinematic equation for uniformly accelerated motion: ?? = ?? 0 + ?? 0?? ?? + 1 2 ?? ?? ?? 2 Given: Page 5 JEE Main Previous Year Questions (2021-2025) Motion 2025 Q1: A particle initially at rest starts moving from reference point ?? = ?? along ?? -axis, with velocity ?? that varies as ?? = ?? v?? ?? /?? . The acceleration of the particle is ???? -?? . [JEE Main 2024 (Online) 1st February Evening Shift] Ans: 8 To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is ?? = 4v?? However, this function gives the velocity as a function of position ?? , not as a function of time ?? . Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate ?? with respect to ?? . The chain rule in this context can be stated as follows: ?? = ???? ???? = ???? ???? · ???? ???? Now, because ???? ?? ?? is the velocity ?? itself and ???? ???? is the derivative of the velocity with respect to ?? , we first find ???? ???? : ?? = 4v?? = 4?? 1 2 Differentiating with respect to ?? , we get: ???? ???? = 4· 1 2 ?? - 1 2 = 2?? - 1 2 = 2 v?? Now, because ?? = 4v?? , we can rewrite v?? as ?? 4 . Using this to replace v?? in our expression for ???? ???? , we get: ???? ???? = 2 v?? = 2 ?? 4 = 8 ?? Now, using the chain rule: ?? = ???? ???? = ???? ???? · ???? ???? = 8 ?? · ?? Simplifying this, the velocity terms cancel out, leaving us with: ?? = 8 ms -2 Thus, the acceleration of the particle is 8 ms -2 . Q2: A particle is moving in one dimension (along ?? axis) under the action of a variable force. It's initial position was ???? ?? right of origin. The variation of its position (?? ) with time (?? ) is given as ?? = -?? ?? ?? + ???? ?? ?? + ???? ?? , where ?? is in ?? and ?? is in ?? .The velocity of the particle when its acceleration becomes zero is ?? /?? . [JEE Main 2024 (Online) 1st February Morning Shift] Ans: 52 Position ( x) : The particle's location on the x-axis at a given time. Velocity (v): The rate of change of position with respect to time (?? = ???? ???? ). Acceleration (a): The rate of change of velocity with respect to time (?? = ???? ???? ) The particle's position is given by: ?? (?? )= -3?? 3 + 18?? 2 + 16?? We need to find the velocity when the acceleration is zero. Find the Velocity (v) and Acceleration (a) Functions: Velocity: ?? (?? )= ???? ???? = -9?? 2 + 36?? + 16 Acceleration: ?? (?? )= ???? ???? = -18?? + 36 Find the Time (t) When Acceleration is Zero: ?? (?? )= 0 -18?? + 36 = 0 ?? = 2 seconds Calculate the Velocity at ?? = 2 seconds: ?? (2)= -9(2) 2 + 36(2)+ 16 ?? (2)= 52 m/s Answer: The velocity of the particle when its acceleration becomes zero is 52 m/s. Q3: The displacement and the increase in the velocity of a moving particle in the time interval of ?? to (?? + ?? )?? are ?????? ?? and ???? ?? /?? , respectively. The distance travelled by the particle in (?? + ?? ) ???? is ?? . [JEE Main 2024 (Online) 30th January Morning Shift] Ans: 175 Considering acceleration is constant ?? = ?? + ???? ?? + 50 = ?? + ?? ? ?? = 50 m/s 2 125 = ???? + 1 2 ?? ?? 2 125 = ?? + ?? 2 ? ?? = 100 m/s ? ?? ?? 4?? = ?? + ?? 2 [2n - 1] = 175 m Q4: A ball rolls off the top of a stairway with horizontal velocity ?? . The steps are ?? .?? ?? high and ?? .?? ?? wide. The minimum velocity ?? with which that ball just hits the step 5 of the stairway will be v?? ???? -?? where ?? = [use ?? = ???? ?? /?? ?? ]. [JEE Main 2024 (Online) 29th January Morning Shift] Ans: 2 Q5: A body falling under gravity covers two points ?? and ?? separated by ???? ?? in ?? ?? . The distance of upper point A from the starting point is ?? (use ?? = ???? ???? -?? ). [JEE Main 2024 (Online) 27th January Evening Shift] Ans: 45 Q6: A particle starts from origin at ?? = ?? with a velocity ?? ??ˆ ?? /?? and moves in ?? - ?? plane under action of a force which produces a constant acceleration of (?? ??ˆ + ?? ??ˆ)?? /?? ?? . If the ?? -coordinate of the particle at that instant is ???? ?? , then the speed of the particle at this time is v?? ?? /?? . The value of ?? is [JEE Main 2024 (Online) 27th January Morning Shift] Ans: 673 To solve for the value of ?? , which represents the square of the speed of the particle at the time its ?? - coordinate is 84 m, we need to first determine the time at which the particle reaches this ?? coordinate, and then use this time to calculate its final velocity in both the ?? and ?? directions. The motion of the particle in the ?? -direction can be described by the kinematic equation for uniformly accelerated motion: ?? = ?? 0 + ?? 0?? ?? + 1 2 ?? ?? ?? 2 Given: ?? 0 = 0 m ?? 0?? = 5 m/s ?? ?? = 3 m/s 2 ?? = 84 m (at which we need to find the speed) Substituting these values into the kinematic equation: 84 m = 0 m+ (5 m/s)?? + 1 2 (3 m/s 2 )?? 2 Simplifying this equation, we get: 0 = 3 2 ?? 2 + 5?? - 84 Using the quadratic formula to solve for ?? : ?? = -?? ± v?? 2 - 4???? 2?? where ?? = 3 2 ,?? = 5, and ?? = -84. ?? = -5 ± v (5) 2 - 4( 3 2 )(-84) 2( 3 2 ) ?? = -5 ± v25+ 504 3 ?? = -5 ± v529 3 ?? = -5 ± 23 3 In this scenario, since we're looking for a time when the particle reaches 84 m, we only consider the positive root because time cannot be negative. ?? = 18 3 ?? = 6 s Now we have the time at which the particle's ?? -coordinate is 84 m. Next, we find final velocities in ?? and ?? directions at ?? = 6 s. The final velocity in the ?? -direction can be found using the formula for velocity with constant acceleration:Read More
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FAQs on Motion in a Straight Line: JEE Mains Previous Year Questions (2021 - 2025) - Physics for JEE Main & Advanced
| 1. What is the concept of motion in a straight line in physics? |  |
Ans.Motion in a straight line refers to the movement of an object along a linear path without changing its direction. This type of motion can be described using fundamental concepts such as displacement, velocity, and acceleration. Displacement is the shortest distance from the initial to the final position, velocity is the rate of change of displacement, and acceleration is the rate of change of velocity over time.
| 2. How do I calculate average velocity and average speed? | |
Ans.Average velocity is calculated by taking the total displacement divided by the total time taken for the motion. It can be expressed as \( \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} \). Average speed, on the other hand, is calculated by dividing the total distance traveled by the total time taken, expressed as \( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \). Average speed is always non-negative, while average velocity can be negative if the object moves in the opposite direction.
| 3. What are the equations of motion for uniformly accelerated motion? | |
Ans.The three main equations of motion for uniformly accelerated linear motion are:
1. \( v = u + at \) (final velocity)
2. \( s = ut + \frac{1}{2}at^2 \) (displacement)
3. \( v^2 = u^2 + 2as \) (relationship between velocity and displacement)
where \( u \) is the initial velocity, \( v \) is the final velocity, \( a \) is the acceleration, \( s \) is the displacement, and \( t \) is the time.
| 4. What is the difference between scalar and vector quantities in motion? | |
Ans.Scalar quantities are those that have only magnitude and no direction, such as speed and distance. Vector quantities, on the other hand, have both magnitude and direction, such as velocity and displacement. This distinction is essential when analyzing motion, as it affects how we add and subtract these quantities in physics problems.
| 5. How can graphical representation help in understanding motion in a straight line? | |
Ans.Graphical representation, such as distance-time and velocity-time graphs, provides a visual way to analyze motion. In a distance-time graph, the slope of the line represents velocity, while in a velocity-time graph, the slope represents acceleration. The area under the velocity-time graph gives the displacement. These graphs help in visualizing concepts such as constant speed, acceleration, and deceleration, making it easier to understand the behavior of moving objects.
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