📘 Motion In One Dimension — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Motion In One Dimension for JEE Main with previous year questions on kinematics, velocity-time graphs, and relative motion. Each solution is explained in a simple, step-by-step manner to boost your score.
Motion In One Dimension — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (7 April Shift 2)
A helicopter flying horizontally with a speed of \(360 \, \text{km/h}\) at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is : (use acceleration due to gravity \(g = 10 \, \text{m/s}^2\) and neglect air resistance)
(1) \(2\sqrt{5} \, \text{km}\) (2) \(4 \, \text{km}\)
(3) \(7.2 \, \text{km}\) (4) \(2\sqrt{2} \, \text{km}\)
(1) \(2\sqrt{5} \, \text{km}\) (2) \(4 \, \text{km}\)
(3) \(7.2 \, \text{km}\) (4) \(2\sqrt{2} \, \text{km}\)
✅ Answer & Explanation:
Correct Option: (4)
- Horizontal velocity \(u = 360 \, \text{km/h} = 100 \, \text{m/s}\).
- Horizontal displacement \(x = u \times t = 100 \times 20 = 2000 \, \text{m} = 2 \, \text{km}\).
- Vertical displacement (altitude) \(y = 2 \, \text{km}\).
- Net displacement \(S = \sqrt{x^2 + y^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \, \text{km}\).
- Horizontal velocity \(u = 360 \, \text{km/h} = 100 \, \text{m/s}\).
- Horizontal displacement \(x = u \times t = 100 \times 20 = 2000 \, \text{m} = 2 \, \text{km}\).
- Vertical displacement (altitude) \(y = 2 \, \text{km}\).
- Net displacement \(S = \sqrt{x^2 + y^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \, \text{km}\).
Q2. JEE Main 2025 (4 April Shift 2)
The displacement \(x\) versus time graph is shown below:
(A) The average velocity during 0 to 3 s is 10 m/s
(B) The average velocity during 3 to 5 s is 0 m/s
(C) The instantaneous velocity at \(t = 2 \, \text{s}\) is 5 m/s
(D) The average velocity during 5 to 7 s and instantaneous velocity at \(t = 6.5 \, \text{s}\) are equal
(E) The average velocity from \(t = 0\) to \(t = 9 \, \text{s}\) is zero.
Choose the correct answer from the options given below:
(1) (A), (D), (E) only (2) (B), (C), (D) only
(3) (B), (D), (E) only (4) (B), (C), (E) only
(B) The average velocity during 3 to 5 s is 0 m/s
(C) The instantaneous velocity at \(t = 2 \, \text{s}\) is 5 m/s
(D) The average velocity during 5 to 7 s and instantaneous velocity at \(t = 6.5 \, \text{s}\) are equal
(E) The average velocity from \(t = 0\) to \(t = 9 \, \text{s}\) is zero.
Choose the correct answer from the options given below:
(1) (A), (D), (E) only (2) (B), (C), (D) only
(3) (B), (D), (E) only (4) (B), (C), (E) only
✅ Answer & Explanation:
Correct Option: (4)
Based on the provided graph:
- (B) Displacement is constant between 3-5s, so velocity is zero.
- (C) Slope at \(t=2s\) is \(\frac{5 - (-5)}{3 - 1} = 5 \, \text{m/s}\).
- (E) Displacement at \(t=0\) and \(t=9s\) is the same, so average velocity is zero.
Based on the provided graph:
- (B) Displacement is constant between 3-5s, so velocity is zero.
- (C) Slope at \(t=2s\) is \(\frac{5 - (-5)}{3 - 1} = 5 \, \text{m/s}\).
- (E) Displacement at \(t=0\) and \(t=9s\) is the same, so average velocity is zero.
Q3. JEE Main 2025 (29 Jan Shift 2)
Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. Both cars cross each other at time \(t = 0\), for the first time. The maximum possible number of crossing(s) (including the crossing at \(t = 0\)) is ________.
✅ Answer & Explanation:
Correct Answer: 3
For car Q (constant \(a\)), position \(x_Q\) is a quadratic function of time. For car P (acceleration \(\propto t\)), position \(x_P\) is a cubic function of time. The equation \(x_P = x_Q\) (representing a crossing) is a cubic equation, which can have a maximum of 3 real roots.
For car Q (constant \(a\)), position \(x_Q\) is a quadratic function of time. For car P (acceleration \(\propto t\)), position \(x_P\) is a cubic function of time. The equation \(x_P = x_Q\) (representing a crossing) is a cubic equation, which can have a maximum of 3 real roots.
Q4. JEE Main 2025 (29 Jan Shift 1)
The maximum speed of a boat in still water is \(27 \, \text{km/h}\). Now this boat is moving downstream in a river flowing at \(9 \, \text{km/h}\). A man in the boat throws a ball vertically upwards with speed of \(10 \, \text{m/s}\). Range of the ball as observed by an observer at rest on the river bank, is _______ cm. (Take \(g = 10 \, \text{m/s}^2\))
✅ Answer & Explanation:
Correct Answer: 2000
- Velocity of boat relative to bank (downstream) \(= 27 + 9 = 36 \, \text{km/h} = 10 \, \text{m/s}\).
- This is the horizontal velocity (\(u_x\)) of the ball.
- Time of flight \(T = \frac{2u_y}{g} = \frac{2 \times 10}{10} = 2 \, \text{s}\).
- Range \(R = u_x \times T = 10 \times 2 = 20 \, \text{m} = 2000 \, \text{cm}\).
- Velocity of boat relative to bank (downstream) \(= 27 + 9 = 36 \, \text{km/h} = 10 \, \text{m/s}\).
- This is the horizontal velocity (\(u_x\)) of the ball.
- Time of flight \(T = \frac{2u_y}{g} = \frac{2 \times 10}{10} = 2 \, \text{s}\).
- Range \(R = u_x \times T = 10 \times 2 = 20 \, \text{m} = 2000 \, \text{cm}\).
Q5. JEE Main 2025 (23 Jan Shift 1)
The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km.
(1) 12 (2) 3
(3) 6 (4) 9
(3) 6 (4) 9
✅ Answer & Explanation:
Correct Option: (1)
Distance is the area under the velocity-time graph. Calculating the total area for the first 30.5 seconds yields 12 km.
Distance is the area under the velocity-time graph. Calculating the total area for the first 30.5 seconds yields 12 km.
Q6. JEE Main 2025 (2 April Shift 1)
A person travelling on a straight line moves with a uniform velocity \(v_1\) for a distance \(x\) and with a uniform velocity \(v_2\) for the next \(\frac{3}{2}x\) distance. The average velocity in this motion is \(\frac{50}{7} \, \text{m/s}\). If \(v_1\) is \(5 \, \text{m/s}\) then \(v_2 = \) _________ m/s.
✅ Answer & Explanation:
Correct Answer: 10
Average velocity \(= \frac{\text{Total Distance}}{\text{Total Time}}\)
\(\frac{x + 1.5x}{x/v_1 + 1.5x/v_2} = \frac{50}{7}\)
\(\frac{2.5}{1/5 + 1.5/v_2} = \frac{50}{7} \implies \frac{1}{5} + \frac{1.5}{v_2} = 0.35 \implies \frac{1.5}{v_2} = 0.15 \implies v_2 = 10 \, \text{m/s}\).
Average velocity \(= \frac{\text{Total Distance}}{\text{Total Time}}\)
\(\frac{x + 1.5x}{x/v_1 + 1.5x/v_2} = \frac{50}{7}\)
\(\frac{2.5}{1/5 + 1.5/v_2} = \frac{50}{7} \implies \frac{1}{5} + \frac{1.5}{v_2} = 0.35 \implies \frac{1.5}{v_2} = 0.15 \implies v_2 = 10 \, \text{m/s}\).
Q7. JEE Main 2024 (31 Jan Shift 1)
The relation between time ‘t’ and distance ‘x’ is \(t = \alpha x^2 + \beta x\), where \(\alpha\) and \(\beta\) are constants. The relation between acceleration (a) and velocity (v) is:
(1) \(a = -2\alpha v^3\) (2) \(a = -5\alpha v^5\)
(3) \(a = -3\alpha v^2\) (4) \(a = -4\alpha v^4\)
(1) \(a = -2\alpha v^3\) (2) \(a = -5\alpha v^5\)
(3) \(a = -3\alpha v^2\) (4) \(a = -4\alpha v^4\)
✅ Answer & Explanation:
Correct Option: (1)
Differentiating wrt \(t\): \(1 = 2\alpha x \frac{dx}{dt} + \beta \frac{dx}{dt} \implies v = \frac{1}{2\alpha x + \beta}\).
Differentiating \(v\) wrt \(t\): \(a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}\).
\(\frac{dv}{dx} = \frac{-2\alpha}{(2\alpha x + \beta)^2} = -2\alpha v^2\).
Thus, \(a = (-2\alpha v^2)(v) = -2\alpha v^3\).
Differentiating wrt \(t\): \(1 = 2\alpha x \frac{dx}{dt} + \beta \frac{dx}{dt} \implies v = \frac{1}{2\alpha x + \beta}\).
Differentiating \(v\) wrt \(t\): \(a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}\).
\(\frac{dv}{dx} = \frac{-2\alpha}{(2\alpha x + \beta)^2} = -2\alpha v^2\).
Thus, \(a = (-2\alpha v^2)(v) = -2\alpha v^3\).
Q8. JEE Main 2024 (27 Jan Shift 2)
A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm. It penetrates further \(D \times 10^{-3} \, \text{m}\) before coming to rest. The value of D is :
(1) 32 (2) 5
(3) 3 (4) 4
(1) 32 (2) 5
(3) 3 (4) 4
✅ Answer & Explanation:
Correct Option: (1)
Using \(v^2 - u^2 = 2as\):
- Case 1: \((\frac{2}{3}u)^2 - u^2 = 2a(0.04) \implies a = \frac{-5u^2}{18(0.04)}\).
- Case 2: \(0 - (\frac{2}{3}u)^2 = 2a(s_2) \implies \frac{4}{9}u^2 = 2(\frac{5u^2}{0.72})s_2 \implies s_2 = 0.032 \, \text{m}\).
Given \(s_2 = D \times 10^{-3} \implies D = 32\).
Using \(v^2 - u^2 = 2as\):
- Case 1: \((\frac{2}{3}u)^2 - u^2 = 2a(0.04) \implies a = \frac{-5u^2}{18(0.04)}\).
- Case 2: \(0 - (\frac{2}{3}u)^2 = 2a(s_2) \implies \frac{4}{9}u^2 = 2(\frac{5u^2}{0.72})s_2 \implies s_2 = 0.032 \, \text{m}\).
Given \(s_2 = D \times 10^{-3} \implies D = 32\).
Q9. JEE Main 2023 (06 Apr Shift 1)
A particle of mass 10 g moves in a straight line with retardation \(2x\), where \(x\) is the displacement in SI units. Its loss of kinetic energy for above displacement is \((\frac{10}{x})^{-n} \, \text{J}\). The value of n will be ________.
✅ Answer & Explanation:
Correct Answer: 2
Loss in KE = Work done against retardation \(= \int F dx = \int ma dx\).
\(\text{Loss} = \int_0^x (0.01)(2x) dx = 0.01 x^2 = \frac{x^2}{100} = (\frac{10}{x})^{-2} \, \text{J}\).
Loss in KE = Work done against retardation \(= \int F dx = \int ma dx\).
\(\text{Loss} = \int_0^x (0.01)(2x) dx = 0.01 x^2 = \frac{x^2}{100} = (\frac{10}{x})^{-2} \, \text{J}\).
Q10. JEE Main 2023 (01 Feb Shift 2)
For a train engine moving with speed of \(20 \, \text{ms}^{-1}\), the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed \(\sqrt{x} \, \text{ms}^{-1}\). The value of \(x\) is ______.
✅ Answer & Explanation:
Correct Answer: 200
- Using \(v^2 = u^2 + 2as \implies 0 = 20^2 + 2a(500) \implies a = -0.4 \, \text{m/s}^2\).
- New distance \(s' = 250 \, \text{m}\).
- \(v'^2 = 20^2 + 2(-0.4)(250) = 400 - 200 = 200\).
- \(v' = \sqrt{200} \implies x = 200\).
- Using \(v^2 = u^2 + 2as \implies 0 = 20^2 + 2a(500) \implies a = -0.4 \, \text{m/s}^2\).
- New distance \(s' = 250 \, \text{m}\).
- \(v'^2 = 20^2 + 2(-0.4)(250) = 400 - 200 = 200\).
- \(v' = \sqrt{200} \implies x = 200\).
Q11. JEE Main 2022 (29 Jul Shift 2)
A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws \(n\) balls per second, the maximum height the balls can reach is:
(1) \(\frac{g}{2n}\) (2) \(\frac{g}{n}\)
(3) \(2gn\) (4) \(\frac{g}{2n^2}\)
(1) \(\frac{g}{2n}\) (2) \(\frac{g}{n}\)
(3) \(2gn\) (4) \(\frac{g}{2n^2}\)
✅ Answer & Explanation:
Correct Option: (4)
- Time between throws \(t = 1/n\).
- At highest point, velocity is zero, so \(t = u/g \implies u = g/n\).
- Max height \(H = \frac{u^2}{2g} = \frac{(g/n)^2}{2g} = \frac{g}{2n^2}\).
- Time between throws \(t = 1/n\).
- At highest point, velocity is zero, so \(t = u/g \implies u = g/n\).
- Max height \(H = \frac{u^2}{2g} = \frac{(g/n)^2}{2g} = \frac{g}{2n^2}\).
Q12. JEE Main 2022 (27 Jul Shift 1)
A bullet is shot vertically downwards with an initial velocity of \(100 \, \text{m/s}\) from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest. The velocity-time curve for total time \(t = 20 \, \text{s}\) will be: (Take \(g = 10 \, \text{m/s}^2\))
Choose the correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (1)
- At \(t=0\), \(v = -100 \, \text{m/s}\) (downward).
- At \(t=10s\), \(v = -100 + (-10)(10) = -200 \, \text{m/s}\).
- After \(t=10s\), the bullet is at rest (\(v=0\)). Graph (1) correctly shows the negative linear increase followed by zero.
- At \(t=0\), \(v = -100 \, \text{m/s}\) (downward).
- At \(t=10s\), \(v = -100 + (-10)(10) = -200 \, \text{m/s}\).
- After \(t=10s\), the bullet is at rest (\(v=0\)). Graph (1) correctly shows the negative linear increase followed by zero.
Q13. JEE Main 2021 (31 Aug Shift 2)
A particle is moving with constant acceleration \(a\). Following graph shows \(v^2\) versus \(x\) (displacement) plot. The acceleration of the particle is _________ \(\text{m/s}^2\).
✅ Answer & Explanation:
Correct Answer: 1
Using \(v^2 = u^2 + 2ax\), the slope of a \(v^2\) vs \(x\) graph is \(2a\).
From the graph, slope \(= \frac{80 - 20}{30 - 0} = \frac{60}{30} = 2\).
So, \(2a = 2 \implies a = 1 \, \text{m/s}^2\).
Using \(v^2 = u^2 + 2ax\), the slope of a \(v^2\) vs \(x\) graph is \(2a\).
From the graph, slope \(= \frac{80 - 20}{30 - 0} = \frac{60}{30} = 2\).
So, \(2a = 2 \implies a = 1 \, \text{m/s}^2\).
Q14. JEE Main 2020 (04 Sep Shift 1)
A tennis ball is released from a height \(h\) and after freely falling on a wooden floor it rebounds and reaches height \(h/2\). The velocity versus height of the ball during its motion may be represented graphically by:
Choose the correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (3)
The relationship between velocity and height is \(v^2 \propto h\), which is parabolic. Graph (3) correctly depicts the downward fall and reduced rebound height with proper direction changes.
The relationship between velocity and height is \(v^2 \propto h\), which is parabolic. Graph (3) correctly depicts the downward fall and reduced rebound height with proper direction changes.
Q15. JEE Main 2020 (02 Sep Shift 1)
Train A and train B are running on parallel tracks in opposite directions with speeds of \(36 \, \text{km/h}\) and \(72 \, \text{km/h}\), respectively. A person is walking in train A in the direction opposite to its motion with a speed of \(1.8 \, \text{km/h}\). Speed (in m/s) of this person as observed from train B will be close to:
(1) 29.5 (2) 28.5
(3) 31.5 (4) 30.5
(1) 29.5 (2) 28.5
(3) 31.5 (4) 30.5
✅ Answer & Explanation:
Correct Option: (1)
- \(V_A = 10 \, \text{m/s}\), \(V_B = -20 \, \text{m/s}\) (opposite directions).
- Velocity of person wrt Train A: \(V_{p,A} = -0.5 \, \text{m/s}\).
- Velocity of person wrt ground: \(V_p = V_A + V_{p,A} = 10 - 0.5 = 9.5 \, \text{m/s}\).
- Velocity of person wrt Train B: \(V_{p,B} = V_p - V_B = 9.5 - (-20) = 29.5 \, \text{m/s}\).
- \(V_A = 10 \, \text{m/s}\), \(V_B = -20 \, \text{m/s}\) (opposite directions).
- Velocity of person wrt Train A: \(V_{p,A} = -0.5 \, \text{m/s}\).
- Velocity of person wrt ground: \(V_p = V_A + V_{p,A} = 10 - 0.5 = 9.5 \, \text{m/s}\).
- Velocity of person wrt Train B: \(V_{p,B} = V_p - V_B = 9.5 - (-20) = 29.5 \, \text{m/s}\).
📘 Exam Preparation Tip for Motion in One Dimension
- Master the Three Equations of Motion: Memorize and understand how to apply \(v = u + at\), \(s = ut + \frac{1}{2}at^2\), and \(v^2 = u^2 + 2as\) for constant acceleration problems.
- Graph Interpretation: Practice deriving displacement (area under \(v\)-\(t\) curve), velocity (slope of \(x\)-\(t\) curve), and acceleration (slope of \(v\)-\(t\) curve) from graphs. This is a common source of questions.
- Relative Motion: Understand the concept of relative velocity and how to apply it to problems involving trains, boats, and airplanes. The key is to define your frames of reference clearly.
- Calculus in Kinematics: Be comfortable with \(v = \frac{dx}{dt}\) and \(a = \frac{dv}{dt}\). For variable acceleration, integration is required to find displacement and velocity.
- Visualize the Problem: Many students find it helpful to draw a diagram or a mental picture of the motion. This helps in correctly assigning signs to displacement, velocity, and acceleration.