Motion in a Straight Line: JEE Main Previous Year Questions (2021-2026)

 Page 1


 
JEE Main Previous Year Questions (2021-2026): 
Motion in a Straight Line  
(January 2026) 
 
 
Q1: A particle starts moving from time t = 0 and its coordinate is given as x (t) = 4t
3
 - 3t  
A. The particle returns to its original position (origin) 0.866 units later  
B. The particle is 1 unit away from origin at its turning point  
C. Acceleration of the particle is non-negative  
D. The particle is 0.5 units away from origin at its turning point  
E. Particle never turns back as acceleration is non-negative  
Choose the correct answer from the options given below :  
(a) A, C, D Only  
(b) A, B, C Only  
(c) C, E Only  
(d) A, C Only 
Ans: (b) 
Solution: 
The original position at t = 0 is x (0) = 4 (0)
3
 - 3 (0) = 0.  
To find when it returns to the origin:  
4t
3
 - 3t = 0  
? t (4t
2
 - 3) = 0  
  
Hence, the Statement A is correct.  
A turning point occurs when velocity v (t) = 0.  
  
At turning point,  
v (t) = 0 ? 12t
2
 = 3 ? t
2
 = 0.25 ? t = 0.5 s  
So, the position at turning point is,  
x (0.5) = 4 (0.5) 3 - 3 (0. ) = 4 (0.125) - 1.5 = 0.5 - 1.5 = - 1.0 m  
The distance from the origin is | - 1.0 | = 1 unit.  
Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect 
because the particle does turn back at t = 0.5 s.  
Acceleration a (t) is the derivative of velocity:  
 
Page 2


 
JEE Main Previous Year Questions (2021-2026): 
Motion in a Straight Line  
(January 2026) 
 
 
Q1: A particle starts moving from time t = 0 and its coordinate is given as x (t) = 4t
3
 - 3t  
A. The particle returns to its original position (origin) 0.866 units later  
B. The particle is 1 unit away from origin at its turning point  
C. Acceleration of the particle is non-negative  
D. The particle is 0.5 units away from origin at its turning point  
E. Particle never turns back as acceleration is non-negative  
Choose the correct answer from the options given below :  
(a) A, C, D Only  
(b) A, B, C Only  
(c) C, E Only  
(d) A, C Only 
Ans: (b) 
Solution: 
The original position at t = 0 is x (0) = 4 (0)
3
 - 3 (0) = 0.  
To find when it returns to the origin:  
4t
3
 - 3t = 0  
? t (4t
2
 - 3) = 0  
  
Hence, the Statement A is correct.  
A turning point occurs when velocity v (t) = 0.  
  
At turning point,  
v (t) = 0 ? 12t
2
 = 3 ? t
2
 = 0.25 ? t = 0.5 s  
So, the position at turning point is,  
x (0.5) = 4 (0.5) 3 - 3 (0. ) = 4 (0.125) - 1.5 = 0.5 - 1.5 = - 1.0 m  
The distance from the origin is | - 1.0 | = 1 unit.  
Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect 
because the particle does turn back at t = 0.5 s.  
Acceleration a (t) is the derivative of velocity:  
 
Since the particle starts moving from t = 0 , for all t = 0 , the value of 24 t will always be greater 
than or equal to zero.  
Hence, statement C is correct. Therefore, statements A , B, and C and D are correct. The option 
2 best represents the correct set.  
Hence, the correct option is (B). 
 
Q2: Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the 
first drop strikes the floor when the sixth drop begins to fall. The height at which the 
fourth drop will be from ground, at the instant when the first drop strikes the ground is 
_____ m.  (g = 10m/s
2
) 
(a) 3.8 
(b) 4.0 
(c) 4.2 
(d) 2.5 
Ans: (c)  
Solution: 
 
The first drop falls a distance of H = 5 m starting from rest (u = 0).  
Page 3


 
JEE Main Previous Year Questions (2021-2026): 
Motion in a Straight Line  
(January 2026) 
 
 
Q1: A particle starts moving from time t = 0 and its coordinate is given as x (t) = 4t
3
 - 3t  
A. The particle returns to its original position (origin) 0.866 units later  
B. The particle is 1 unit away from origin at its turning point  
C. Acceleration of the particle is non-negative  
D. The particle is 0.5 units away from origin at its turning point  
E. Particle never turns back as acceleration is non-negative  
Choose the correct answer from the options given below :  
(a) A, C, D Only  
(b) A, B, C Only  
(c) C, E Only  
(d) A, C Only 
Ans: (b) 
Solution: 
The original position at t = 0 is x (0) = 4 (0)
3
 - 3 (0) = 0.  
To find when it returns to the origin:  
4t
3
 - 3t = 0  
? t (4t
2
 - 3) = 0  
  
Hence, the Statement A is correct.  
A turning point occurs when velocity v (t) = 0.  
  
At turning point,  
v (t) = 0 ? 12t
2
 = 3 ? t
2
 = 0.25 ? t = 0.5 s  
So, the position at turning point is,  
x (0.5) = 4 (0.5) 3 - 3 (0. ) = 4 (0.125) - 1.5 = 0.5 - 1.5 = - 1.0 m  
The distance from the origin is | - 1.0 | = 1 unit.  
Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect 
because the particle does turn back at t = 0.5 s.  
Acceleration a (t) is the derivative of velocity:  
 
Since the particle starts moving from t = 0 , for all t = 0 , the value of 24 t will always be greater 
than or equal to zero.  
Hence, statement C is correct. Therefore, statements A , B, and C and D are correct. The option 
2 best represents the correct set.  
Hence, the correct option is (B). 
 
Q2: Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the 
first drop strikes the floor when the sixth drop begins to fall. The height at which the 
fourth drop will be from ground, at the instant when the first drop strikes the ground is 
_____ m.  (g = 10m/s
2
) 
(a) 3.8 
(b) 4.0 
(c) 4.2 
(d) 2.5 
Ans: (c)  
Solution: 
 
The first drop falls a distance of H = 5 m starting from rest (u = 0).  
Using the equation of motion: 
 
When the first drop hits the ground, the sixth drop is just beginning to fall. This means there are 
5 equal time intervals between the first and sixth drops. 
 
At the instant the first drop strikes the ground (t = 1s): 
Drop 1 has fallen for 1.0 s 
Drop 2 has fallen for 0.8 s 
Drop 3 has fallen for 0.6 s 
Drop 4 has fallen for 0.4 s 
Using the distance formula for the fourth drop: 
 
The height from the ground is the total height minus the distance fallen: 
Height from ground = H - h
4
 = 5 m - 0.8 m = 4.2 m  
Hence, the correct option is (C). 
 
Q3: The velocity (v) - Distance (x) graph is shown in figure. Which graph represents 
acceleration(a) versus distance (x) variation of this system? 
Page 4


 
JEE Main Previous Year Questions (2021-2026): 
Motion in a Straight Line  
(January 2026) 
 
 
Q1: A particle starts moving from time t = 0 and its coordinate is given as x (t) = 4t
3
 - 3t  
A. The particle returns to its original position (origin) 0.866 units later  
B. The particle is 1 unit away from origin at its turning point  
C. Acceleration of the particle is non-negative  
D. The particle is 0.5 units away from origin at its turning point  
E. Particle never turns back as acceleration is non-negative  
Choose the correct answer from the options given below :  
(a) A, C, D Only  
(b) A, B, C Only  
(c) C, E Only  
(d) A, C Only 
Ans: (b) 
Solution: 
The original position at t = 0 is x (0) = 4 (0)
3
 - 3 (0) = 0.  
To find when it returns to the origin:  
4t
3
 - 3t = 0  
? t (4t
2
 - 3) = 0  
  
Hence, the Statement A is correct.  
A turning point occurs when velocity v (t) = 0.  
  
At turning point,  
v (t) = 0 ? 12t
2
 = 3 ? t
2
 = 0.25 ? t = 0.5 s  
So, the position at turning point is,  
x (0.5) = 4 (0.5) 3 - 3 (0. ) = 4 (0.125) - 1.5 = 0.5 - 1.5 = - 1.0 m  
The distance from the origin is | - 1.0 | = 1 unit.  
Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect 
because the particle does turn back at t = 0.5 s.  
Acceleration a (t) is the derivative of velocity:  
 
Since the particle starts moving from t = 0 , for all t = 0 , the value of 24 t will always be greater 
than or equal to zero.  
Hence, statement C is correct. Therefore, statements A , B, and C and D are correct. The option 
2 best represents the correct set.  
Hence, the correct option is (B). 
 
Q2: Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the 
first drop strikes the floor when the sixth drop begins to fall. The height at which the 
fourth drop will be from ground, at the instant when the first drop strikes the ground is 
_____ m.  (g = 10m/s
2
) 
(a) 3.8 
(b) 4.0 
(c) 4.2 
(d) 2.5 
Ans: (c)  
Solution: 
 
The first drop falls a distance of H = 5 m starting from rest (u = 0).  
Using the equation of motion: 
 
When the first drop hits the ground, the sixth drop is just beginning to fall. This means there are 
5 equal time intervals between the first and sixth drops. 
 
At the instant the first drop strikes the ground (t = 1s): 
Drop 1 has fallen for 1.0 s 
Drop 2 has fallen for 0.8 s 
Drop 3 has fallen for 0.6 s 
Drop 4 has fallen for 0.4 s 
Using the distance formula for the fourth drop: 
 
The height from the ground is the total height minus the distance fallen: 
Height from ground = H - h
4
 = 5 m - 0.8 m = 4.2 m  
Hence, the correct option is (C). 
 
Q3: The velocity (v) - Distance (x) graph is shown in figure. Which graph represents 
acceleration(a) versus distance (x) variation of this system? 
 
(a)  
 
Page 5


 
JEE Main Previous Year Questions (2021-2026): 
Motion in a Straight Line  
(January 2026) 
 
 
Q1: A particle starts moving from time t = 0 and its coordinate is given as x (t) = 4t
3
 - 3t  
A. The particle returns to its original position (origin) 0.866 units later  
B. The particle is 1 unit away from origin at its turning point  
C. Acceleration of the particle is non-negative  
D. The particle is 0.5 units away from origin at its turning point  
E. Particle never turns back as acceleration is non-negative  
Choose the correct answer from the options given below :  
(a) A, C, D Only  
(b) A, B, C Only  
(c) C, E Only  
(d) A, C Only 
Ans: (b) 
Solution: 
The original position at t = 0 is x (0) = 4 (0)
3
 - 3 (0) = 0.  
To find when it returns to the origin:  
4t
3
 - 3t = 0  
? t (4t
2
 - 3) = 0  
  
Hence, the Statement A is correct.  
A turning point occurs when velocity v (t) = 0.  
  
At turning point,  
v (t) = 0 ? 12t
2
 = 3 ? t
2
 = 0.25 ? t = 0.5 s  
So, the position at turning point is,  
x (0.5) = 4 (0.5) 3 - 3 (0. ) = 4 (0.125) - 1.5 = 0.5 - 1.5 = - 1.0 m  
The distance from the origin is | - 1.0 | = 1 unit.  
Hence, the statements B is correct and statement D is incorrect. Statement E is incorrect 
because the particle does turn back at t = 0.5 s.  
Acceleration a (t) is the derivative of velocity:  
 
Since the particle starts moving from t = 0 , for all t = 0 , the value of 24 t will always be greater 
than or equal to zero.  
Hence, statement C is correct. Therefore, statements A , B, and C and D are correct. The option 
2 best represents the correct set.  
Hence, the correct option is (B). 
 
Q2: Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the 
first drop strikes the floor when the sixth drop begins to fall. The height at which the 
fourth drop will be from ground, at the instant when the first drop strikes the ground is 
_____ m.  (g = 10m/s
2
) 
(a) 3.8 
(b) 4.0 
(c) 4.2 
(d) 2.5 
Ans: (c)  
Solution: 
 
The first drop falls a distance of H = 5 m starting from rest (u = 0).  
Using the equation of motion: 
 
When the first drop hits the ground, the sixth drop is just beginning to fall. This means there are 
5 equal time intervals between the first and sixth drops. 
 
At the instant the first drop strikes the ground (t = 1s): 
Drop 1 has fallen for 1.0 s 
Drop 2 has fallen for 0.8 s 
Drop 3 has fallen for 0.6 s 
Drop 4 has fallen for 0.4 s 
Using the distance formula for the fourth drop: 
 
The height from the ground is the total height minus the distance fallen: 
Height from ground = H - h
4
 = 5 m - 0.8 m = 4.2 m  
Hence, the correct option is (C). 
 
Q3: The velocity (v) - Distance (x) graph is shown in figure. Which graph represents 
acceleration(a) versus distance (x) variation of this system? 
 
(a)  
 
(b)  
 
(c)  
 
(d)