Equations of Motion, Motion Under Gravity, Variable Acceleration, and v-t, x-t, a-t Graphs
IIT-JEE MAINS PYQ Arena — Set 25
Equations of Motion, Gravity, Variable Acceleration & Kinematic Graphs
Set 25 • 375-PYQ Series
Equations of Motion (Consecutive Train Carriages Passing Observer)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. A passenger standing on a railway platform stands right next to the front end of the 1st bogie (carriage) of a stationary train. At t = 0, the train starts moving with a constant acceleration. If the 1st bogie takes 4 seconds to completely pass the passenger, how much time does the 2nd identical bogie take to pass him, and how many complete bogies pass him in the first 12 seconds?
✔ Cumulative Bogie Law N = (t / t1)2 Mastered! +4 Marks
✖ For 1 bogie, L = (1/2) a t12; for N bogies, N L = (1/2) a tN2 ⇒ tN = t1 √N!
Concept & Solution:• Since the train starts from rest, time for N bogies of length L each to pass is tN = t1 √N.
• Time for the 2nd bogie alone: Δt2 = t2 − t1 = t1 (√2 − 1) = 4(√2 − 1) s.
• Number of bogies passing in t = 12 s: N = (t / t1)2 = (12 / 4)2 = 9 bogies.
Question 1 of 15
Motion Under Gravity (Symmetric Speeds Above & Below Point P)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A stone is projected vertically upwards from a platform at point P. When the stone reaches a point A at a distance h = 15 m below P (on its way down), its speed is found to be double the speed it had when it was at a point B at the exact same distance h = 15 m above P. Taking g = 10 m s−2, what is the maximum height (Hmax) reached by the stone above the projection point P?
✔ Symmetric Displacement Equation vA2 = 4 vB2 ⇒ Hmax = (5/3)h Mastered! +4 Marks
✖ At B (+h), vB2 = u2 − 2gh; at A (−h), vA2 = u2 + 2gh = 4(u2 − 2gh) ⇒ 3u2 = 10gh ⇒ Hmax = u2/(2g) = (5/3)h = 25 m!
Concept & Solution:• At point B (distance h above P): vB2 = u2 − 2 g h.
• At point A (distance h below P): vA2 = u2 + 2 g h.
• Given vA = 2 vB ⇒ vA2 = 4 vB2:
• u2 + 2 g h = 4 (u2 − 2 g h) ⇒ 3 u2 = 10 g h ⇒ Hmax = u2 / (2 g) = (5 / 3) h = (5 / 3) × 15 m = 25 m.
x-t Graph Tangent Angles & Velocity Ratio
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. On the position-time (x-t) graph shown below, two straight lines representing the uniform motion of particles P and Q make angles of 30° with the time axis and 30° with the position (vertical) axis respectively. What is the ratio of the velocity of particle P to that of particle Q (vP / vQ)?
✔ Vertical Axis Angle Trap Caught! (θQ = 90° − 30° = 60° with t-axis) +4 Marks
✖ Watch the axis! Line Q makes 30° with the VERTICAL position axis, so it makes 60° with the time axis: vP/vQ = tan 30° / tan 60° = 1/3!
Concept & Solution:• Velocity is the slope dx/dt = tan θ, where θ is always measured from the horizontal time (t) axis.
• For particle P: θP = 30° ⇒ vP = tan 30° = 1 / √3.
• For particle Q: 30° with vertical x-axis means θQ = 90° − 30° = 60° with the t-axis ⇒ vQ = tan 60° = √3.
• Ratio vP / vQ = (1 / √3) / √3 = 1 : 3.
Variable Acceleration (Hyperbolic Velocity v = k / x)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A particle moves along the positive x-axis such that its instantaneous velocity is inversely proportional to its position: v(x) = 12 / x (in SI units). If the particle is at x = 2 m at t = 0, what are its position x and instantaneous acceleration a at t = 1 s?
✔ Separation x dx = 12 dt & Chain Rule a = −k2 / x3 Mastered! +4 Marks
✖ Integrate x dx = 12 dt from x = 2 to x: (x2 − 4)/2 = 12(1) ⇒ x = √28 = 2√7 m, and a = v(dv/dx) = −144 / x3!
Concept & Solution:• Since dx/dt = 12 / x ⇒ x dx = 12 dt.
• Integrating from (t = 0, x = 2 m) to (t = 1 s, x): (x2 − 22) / 2 = 12 × 1 ⇒ x2 = 28 ⇒ x = √28 = 2√7 m.
• Instantaneous acceleration: a = v (dv/dx) = (12 / x) × (−12 / x2) = −144 / x3.
• At x = 2√7 m: a = −144 / (8 × 7√7) = −18 / (7√7) m s−2 (or with k = 12, −144 / 56√7 = −18 / (7√7) m s−2; if v = 6/x from x = 2 at t = 2 s, x = 2√7 and a = −36 / (56√7) = −9 / (14√7)).
Motion Under Gravity (Two Stones Dropped from Different Heights)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. Stone A is dropped from rest from a height of H1 = 80 m. Exactly 1 second later, Stone B is dropped from rest from a lower balcony at height H2 above the ground. If both stones hit the ground at the exact same instant, what is the height H2 of the balcony? (Take g = 10 m s−2)
✔ Fall Time Difference t2 = t1 − 1 = 3 s → 45 m Nailed! +4 Marks
✖ Stone A takes t1 = √(2×80/10) = 4 s to hit the ground! Since Stone B is dropped 1 s later, it falls for t2 = 3 s, covering H2 = 5(32) = 45 m!
Concept & Solution:• Time of fall for Stone A: t1 = √(2 H1 / g) = √(160 / 10) = 4 s.
• Since Stone B is released 1 s later and lands simultaneously, its time of fall is t2 = 4 s − 1 s = 3 s.
• Height of balcony: H2 = (1 / 2) g t22 = 0.5 × 10 × (3)2 = 45 m.
Equations of Motion (α-Cruising-β Three-Stage Journey Time)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. An electric car starts from rest, accelerates uniformly at α = 2 m s−2 until it reaches a top speed of vm = 20 m s−1, maintains this constant top speed for some distance, and finally decelerates uniformly at β = 4 m s−2 to come to rest at a destination S = 550 m away from the starting point. What is the total time (T) taken for the 550 m trip?
✔ Three-Stage Time Formula T = S/vm + (vm/2)(1/α + 1/β) Mastered! +4 Marks
✖ t1 = 20/2 = 10 s (s1 = 100 m) and t3 = 20/4 = 5 s (s3 = 50 m). Remaining 400 m at 20 m s−1 takes t2 = 20 s, so T = 10 + 20 + 5 = 35 s!
Concept & Solution:• Phase 1 (Acceleration): t1 = vm / α = 20 / 2 = 10 s; distance s1 = (1/2)(20)(10) = 100 m.
• Phase 3 (Deceleration): t3 = vm / β = 20 / 4 = 5 s; distance s3 = (1/2)(20)(5) = 50 m.
• Phase 2 (Constant speed for remaining s2 = 550 − 150 = 400 m): t2 = 400 / 20 = 20 s.
• Total Time: T = t1 + t2 + t3 = 10 + 20 + 5 = 35.0 s.
Triangular a-t Graph to Maximum Velocity & Displacement
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A particle starts from rest at x = 0 at t = 0. Its acceleration-time (a-t) graph increases linearly from 0 at t = 0 to a peak of 6 m s−2 at t = 2 s, and then drops vertically to 0 for t > 2 s, as shown below. What are the velocity v(2) and the position x(2) of the particle at t = 2 s?
✔ Linear Ramp a(t) = 3t ⇒ v(t) = 1.5t2 ⇒ x(t) = 0.5t3 Mastered! +4 Marks
✖ Don’t use x = (1/2) v(2) t = 6 m! Since a(t) = 3t increases linearly, v(t) = 1.5t2 is a parabola, so x(2) = 0.5(23) = 4 m!
Concept & Solution:• From t = 0 to t = 2 s, the equation of the acceleration line is a(t) = 3 t m s−2.
• Integrating once (or taking area of triangle): v(t) = (3 / 2) t2 ⇒ at t = 2 s, v(2) = 1.5 × 4 = 6 m s−1.
• Integrating v(t) from 0 to 2 s: x(t) = (1 / 2) t3 ⇒ at t = 2 s, x(2) = 0.5 × (2)3 = 4 m.
Motion Under Gravity (Sound of Splash in a Deep Well)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. A stone is dropped from rest into a vertical well of water depth H = 45 m. If the speed of sound in air is vs = 300 m s−1 and g = 10 m s−2, after how much total time (T) from the instant of dropping the stone is the sound of the splash heard at the top of the well?
✔ Free-Fall Time √(2H/g) + Sound Travel Time H/vs Nailed! +4 Marks
✖ Total time is the sum of stone fall time t1 = √(90/10) = 3.00 s and sound travel time t2 = 45/300 = 0.15 s!
Concept & Solution:• Time taken by the stone to fall 45 m to the water surface:
• t1 = √(2 H / g) = √(2 × 45 / 10) = √9 = 3.00 s.
• Time taken by the sound wave to travel 45 m up the well at constant speed:
• t2 = H / vs = 45 m / 300 m s−1 = 0.15 s.
• Total Time: T = t1 + t2 = 3.00 + 0.15 = 3.15 s.
Equations of Motion (Three Positions in AP vs. GP)
JEE Main 2022 (June Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. A particle moving with uniform acceleration a along a straight line has velocities v1, v2, and v3 at three instants of time t1, t2, and t3 respectively. If the time intervals satisfy (t2 − t1) : (t3 − t2) = 1 : 2, what is the ratio of the average velocity in the first interval to the average velocity in the second interval?
✔ Uniform Acceleration Average Velocity = (vinitial + vfinal)/2 Mastered! +4 Marks
✖ For ANY time interval under constant acceleration, average velocity is simply the arithmetic mean of the endpoint velocities: (vi + vf)/2!
Concept & Solution:• For uniform acceleration, the average velocity over any interval [ti, tf] is always (vi + vf) / 2, regardless of the duration of the interval!
• Average velocity in [t1, t2]: vavg,1 = (v1 + v2) / 2.
• Average velocity in [t2, t3]: vavg,2 = (v2 + v3) / 2.
• Ratio = (v1 + v2) / (v2 + v3) (whereas (v2 − v1)/(v3 − v2) = (a Δt1)/(a Δt2) = 1 : 2).
Variable Acceleration (v(x) = α x−1/2 Time Dependence)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A particle starts from x = 0 at t = 0 and moves along the positive x-axis such that its instantaneous velocity varies with position x as v(x) = α x−1/2, where α is a positive constant. How does the instantaneous velocity v(t) of the particle depend on time t?
✔ Separation x1/2 dx = α dt ⇒ x ∝ t2/3 ⇒ v ∝ t−1/3 Mastered! +4 Marks
✖ Since dx/dt = α x−1/2 ⇒ x1/2 dx = α dt ⇒ x3/2 ∝ t ⇒ x ∝ t2/3, differentiating gives v = dx/dt ∝ t−1/3!
Concept & Solution:• Separating variables in dx/dt = α x−1/2:
• x1/2 dx = α dt ⇒ integrating gives (2 / 3) x3/2 = α t.
• Thus position scales as: x(t) ∝ t2/3.
• Differentiating with respect to t gives velocity: v(t) = dx/dt ∝ t−1/3 (and acceleration a(t) ∝ t−4/3).
Motion Under Gravity (Infinite Geometric Series of Bouncing Ball)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A rubber ball is dropped from rest from an initial height of H = 15 m onto a horizontal floor. On every bounce, the ball loses half of its mechanical energy (so each rebound height is half of the preceding height: hn = H / 2n). What are the Total Distance travelled by the ball and its Net Displacement by the time it finally comes to rest after infinitely many bounces?
✔ Infinite Bounce Series D = H (1 + e2)/(1 − e2) = 3H Mastered! +4 Marks
✖ Every rebound is traversed TWICE (up and down): D = H + 2(H/2 + H/4 + H/8 + …) = H + 2H = 3H = 45 m!
Concept & Solution:• Initial fall = H, followed by up-and-down rebounds of H/2, H/4, H/8, …:
• Total Distance = H + 2 H (1/2 + 1/4 + 1/8 + …) = H + 2 H [ (1/2) / (1 − 1/2) ] = H + 2 H = 3 H = 45 m.
• Since the ball starts at height H = 15 m and ends on the floor (y = 0), Net Displacement = 15 m downwards.
v-t Graph Intersection (Two Cars Overtaking Time)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. Two cars P and Q start from the exact same point at t = 0 and move in the same direction along a straight road. Their velocity-time (v-t) graphs are shown below: Car P moves at a constant velocity of 15 m s−1, while Car Q starts from rest with a constant acceleration such that the two v-t lines intersect at t = 6 s. At what time t does Car Q actually overtake Car P, and what is the maximum lead distance that Car P achieves ahead of Car Q before being overtaken?
✔ v-t Intersection = Maximum Separation (t = 6 s) & Equal Area = Overtaking (t = 12 s)! +4 Marks
✖ Classic JEE Trap! On a v-t graph, intersection at t = 6 s means equal speeds (maximum lead of P = area of triangle = 45 m), while overtaking occurs when areas are equal at t = 12 s!
Concept & Solution:• For t < 6 s, vP > vQ, so Car P pulls ahead of Car Q until their speeds become equal at t = 6 s.
• Maximum Lead of P (at t = 6 s) = Area between the two v-t lines from 0 to 6 s = (1 / 2) × 6 s × 15 m s−1 = 45 m.
• Overtaking occurs when the total areas under both v-t graphs are equal: 15 t = (1 / 2) (2.5 t) t ⇒ t = 12 s.
Equations of Motion (Reaction Time from Two Braking Trials)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A particle moving with uniform retardation along a straight line covers 20 m in the first 2 seconds and 4 m in the next 2 seconds (from t = 2 s to t = 4 s) before coming to rest. What are the initial velocity (u) and the magnitude of retardation (|a|), and at what exact time does it come to rest?
✔ Consecutive Interval Difference S2 − S1 = a T2 Mastered! +4 Marks
✖ Using S2 − S1 = a T2 for T = 2 s: 4 − 20 = 4a ⇒ a = −4 m s−2! Then 20 = 2u − 0.5(4)(4) ⇒ u = 14 m s−1!
Concept & Solution:• Assuming motion continues without reversing before t = 4 s (or checking stopping time):
• S2 − S1 = a T2 ⇒ 4 − 20 = a (2)2 ⇒ a = −4 m s−2.
• From S1 = u(2) + (1/2)(−4)(2)2 = 20 ⇒ 2 u − 8 = 20 ⇒ u = 14 m s−1.
• Stopping time: tstop = u / |a| = 14 / 4 = 3.5 s.
Motion Under Gravity (Upward Throw From Tower Ratio of Times)
JEE Main 2026 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. A stone is thrown vertically upwards from the top of a tower of height H with initial speed u. It reaches the ground in time t1 = n (2 u / g), where (2 u / g) is the time it takes to return to the top of the tower on its way down. Express the height of the tower H in terms of the maximum height above the tower h0 = u2 / (2 g) and the dimensionless factor n (where n > 1):
✔ Tower-to-Apex Ratio H = 4 n(n − 1) h0 Mastered! +4 Marks
✖ Substitute t1 = 2nu/g into −H = u t1 − (1/2)g t12 to get H = (2n2 − 2n)(u2/g) = 4n(n − 1) h0!
Concept & Solution:• Using the vertical displacement equation from the top of the tower to the ground:
• −H = u t1 − (1 / 2) g t12.
• Substituting t1 = 2 n u / g:
• −H = u (2 n u / g) − (1 / 2) g (4 n2 u2 / g2) = (2 n − 2 n2) (u2 / g).
• Since u2 / g = 2 h0, we get H = 4 n (n − 1) h0.
Set 25 Grand Finale (v-x Graph Subnormal & Acceleration)
JEE Main 2026 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. On a velocity-position (v-x) graph for a particle moving along the x-axis, what geometrical quantity at any point P(x, v) on the curve is numerically equal to the instantaneous acceleration (a = v dv/dx) of the particle? And if the v-x curve is the parabola x = (1 / 4) v2, what is the acceleration of the particle?
✔ 375th Question Milestone Conquered! Subnormal of v-x Curve = v(dv/dx) = a! +4 Marks
✖ In coordinate geometry, y(dy/dx) is the SUBNORMAL! For x = v2/4 ⇒ v2 = 4x, comparing with v2 = 2ax gives constant acceleration a = 2 m s−2!
Concept & Solution:• On a v vs. x graph (with v on the vertical y-axis and x on the horizontal axis), the quantity v (dv/dx) is the geometrical Subnormal (projection of the normal segment on the x-axis).
• For the parabola x = (1 / 4) v2 ⇒ v2 = 4 x:
• Differentiating with respect to x gives: 2 v (dv/dx) = 4 ⇒ a = v (dv/dx) = 2 m s−2 (a parabola with constant subnormal!).
Set 25 — 375-PYQ Series Completed!
Phenomenal achievement! You now have 375 unique, zero-dependency JEE Main PYQs across Sets 1–25. Check your Set 25 score below or click any question pill above to review.
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