Motion in a Straight Line: JEE Mains Previous Year Questions (2021-2024) | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Motion 
2024 
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 2


JEE Mains Previous Year Questions 
(2021-2024): Motion 
2024 
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 3


JEE Mains Previous Year Questions 
(2021-2024): Motion 
2024 
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 4


JEE Mains Previous Year Questions 
(2021-2024): Motion 
2024 
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: 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Motion 
2024 
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: 
?? ?? = ?? 0?? + ?? ?? ?? ?? ?? = 5 m/s+ (3 m/s
2
)(6 s)
?? ?? = 5 m/s+ 18 m/s
?? ?? = 23 m/s
 
Similarly, for the ?? -direction: 
?? ?? = ?? 0?? + ?? ?? ?? 
Since the particle starts from the origin and is only subject to a force after ?? = 0, its initial velocity in the 
?? -direction is 0 . 
?? ?? = 0+ (
2 m
s
2
)(6 s) 
?? ?? = 12 m/s 
Now we can compute the speed of the particle, which is the magnitude of the velocity vector: 
?? = v?? ?? 2
+ ?? ?? 2
?? = v(23 m/s)
2
+ (12 m/s)
2
?? = v529+ 144
?? = v673
 
Therefore, the speed of the particle at the time its ?? -coordinate is 84 m is v673 m/s. 
So, ?? = 673. 
Q7: Train A is moving along two parallel rail tracks towards north with speed ???? ???? /?? and train B is 
moving towards south with speed ?????? ???? /?? . Velocity of train B with respect to ?? and velocity of 
ground with respect to ?? are (in ????
-?? ): 
(a) -50 and -30 
(b) -50 and 30 
(c) -30 and 50 
(d) 50 and -30    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (b) 
To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from 
the velocity of Train B, keeping in mind that they are moving in opposite directions. Since they are 
moving in opposite directions, the relative velocity is calculated by adding their magnitudes when 
converting into the same unit, which in this case is meters per second. 
First, let's convert the speeds from km/h to m/s by multiplying by the conversion factor 
 
1000 m/km
3600 s/h
=
5
18
 m/s.