TOPIC: Equations of Motion, Motion Under Gravity, Variable Acceleration, and v-t, x-t, a-t Graphs
IIT-JEE MAINS PYQ Arena — Set 28
Equations of Motion, Gravity, Variable Acceleration & Kinematic Graphs
Set 28 • 420-PYQ Series
Equations of Motion (Time Midpoint vs. Space Midpoint Ratio)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. A particle starts from rest at point A and moves with uniform acceleration along a straight line to reach point B in total time T, covering a total distance S. What is the ratio of its instantaneous speed at the time midpoint (t = T / 2) to its instantaneous speed at the spatial midpoint (x = S / 2)?
✔ Time-Midpoint (vB/2) vs. Space-Midpoint (vB/√2) Ratio Nailed! +4 Marks
✖ Starting from rest, speed at t = T/2 is vB/2, whereas speed at x = S/2 is vB/√2, giving ratio (1/2) / (1/√2) = 1 : √2!
Concept & Solution:• Let the final speed at point B be vB = a T = √(2 a S).
• At the time midpoint (t = T / 2): vtime = a (T / 2) = vB / 2.
• At the spatial midpoint (x = S / 2): vspace = √[ 2 a (S / 2) ] = vB / √2.
• Ratio vtime / vspace = (vB / 2) / (vB / √2) = 1 : √2.
Question 1 of 15
Motion Under Gravity (Successive Drops Separation in Air)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. Two balls A and B are dropped from rest from the roof of a high building at an interval of Δt = 1 s (Ball A is dropped at t = 0, and Ball B is dropped at t = 1 s). Taking g = 10 m s−2, what are the vertical separation and the relative speed between the two balls at t = 3 s after Ball A was dropped?
✔ Constant Relative Velocity vrel = g Δt = 10 m s−1 Mastered! +4 Marks
✖ At t = 3 s, Ball A has fallen for 3 s (yA = 45 m, vA = 30 m s−1) and Ball B has fallen for 2 s (yB = 20 m, vB = 20 m s−1)!
Concept & Solution:• At t = 3 s, Ball A has fallen for 3 s: yA = (1/2)(10)(3)2 = 45 m, and vA = 30 m s−1.
• At t = 3 s, Ball B has fallen for 3 − 1 = 2 s: yB = (1/2)(10)(2)2 = 20 m, and vB = 20 m s−1.
• Vertical Separation = yA − yB = 45 − 20 = 25 m.
• Relative Speed = vA − vB = g Δt = 10 m s−1 (remains constant once both are in free fall!).
a-t Graph with Asymmetrical Positive & Negative Pulses
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A particle is at rest at the origin (x = 0, v = 0) at t = 0. Its acceleration-time (a-t) graph is shown below: a = +4 m s−2 from t = 0 to t = 3 s, and a = −6 m s−2 from t = 3 s to t = 5 s. What are the velocity v(5) and the total displacement x(5) of the particle at t = 5 s?
✔ Triangular v-t Peak (12 m s−1) × Base (5 s) / 2 = 30 m Mastered! +4 Marks
✖ Convert a-t to a v-t triangle! Velocity rises linearly from 0 to +12 m s−1 at t = 3 s, then drops linearly back to 0 at t = 5 s, so Area = 0.5 × 5 s × 12 m s−1 = 30 m!
Concept & Solution:• At t = 3 s: v(3) = 0 + 4 × 3 = +12 m s−1.
• At t = 5 s: v(5) = 12 + (−6) × (5 − 3) = 12 − 12 = 0 m s−1.
• Thus the corresponding v-t graph is a single positive triangle of base 5 s and height 12 m s−1:
• Total Displacement x(5) = (1 / 2) × Base × Height = 0.5 × 5 s × 12 m s−1 = 30 m (18 m in the first 3 s + 12 m in the next 2 s).
Variable Acceleration (a(t) = k / (t + 1)2 Integration)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A particle starts from rest at the origin (x = 0, v = 0) at t = 0 and moves along the positive x-axis with a time-dependent acceleration given by a(t) = 8 / (t + 1)2 m s−2, where t is in seconds. What are its terminal velocity (as t → ∞) and its instantaneous velocity at t = 3 s?
✔ Definite Integral ∫₀ᵗ 8(t + 1)−2 dt = 8[1 − 1/(t + 1)] Mastered! +4 Marks
✖ Integrate a(t) from 0 to t: v(t) = [−8/(t + 1)]₀ᵗ = 8(1 − 1/(t + 1)), giving v(3) = 8(1 − 1/4) = 6 m s−1 and v∞ = 8 m s−1!
Concept & Solution:• Integrating acceleration from 0 to t (with v(0) = 0):
• v(t) = ∫0t 8 (t + 1)−2 dt = [ −8 / (t + 1) ]0t = 8 [ 1 − 1 / (t + 1) ] m s−1.
• As t → ∞: 1 / (t + 1) → 0 ⇒ v∞ = 8 m s−1.
• At t = 3 s: v(3) = 8 (1 − 1 / 4) = 8 × (3 / 4) = 6 m s−1.
Equations of Motion (Three Equal Spatial Segments Time Ratio)
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. A car moving with uniform acceleration along a straight road passes three consecutive milestones A, B, and C that are equally spaced by d = 60 m each (so AB = BC = 60 m). If the car takes t1 = 4 s to travel from A to B and t2 = 2 s to travel from B to C, what are the constant acceleration (a) of the car and its instantaneous speed (vB) as it passes the middle milestone B?
✔ Mid-Time Velocity Shortcut (vavg,BC − vavg,AB) / [(t1 + t2)/2] Mastered! +4 Marks
✖ Average speed in AB is 60/4 = 15 m s−1 (at t = 2 s) and in BC is 60/2 = 30 m s−1 (at t = 5 s), separated by Δt = 3 s, so a = (30 − 15)/3 = 5 m s−2!
Concept & Solution:• Average velocity in AB (equals instantaneous velocity at midpoint of t1, i.e., 2 s before B): v1 = 60 m / 4 s = 15 m s−1.
• Average velocity in BC (equals instantaneous velocity at midpoint of t2, i.e., 1 s after B): v2 = 60 m / 2 s = 30 m s−1.
• Time elapsed between these two mid-time instants: Δt = (t1 + t2) / 2 = (4 + 2) / 2 = 3 s.
• Acceleration: a = (v2 − v1) / Δt = (30 − 15) / 3 = 5 m s−2.
• Speed at milestone B (2 s after v1): vB = 15 + 5(2) = 25 m s−1.
Motion Under Gravity (Gravitational Field Ratio of Two Planets)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. A stone is dropped from a height H on Planet X (where acceleration due to gravity is gX) and reaches the surface of Planet X in time TX with speed vX. When the same stone is dropped from the same height H on Planet Y (where gY = gX / 4), it takes time TY = TX + 2 s and reaches the surface with speed vY = vX − 10 m s−1. What are the values of gX and H?
✔ √g Ratio Scaling (TY = 2 TX, vY = vX/2) Mastered! +4 Marks
✖ Since gY = gX/4, time doubles (TY = 2TX = TX + 2 ⇒ TX = 2 s) and impact speed halves (vY = vX/2 = vX − 10 ⇒ vX = 20 m s−1)!
Concept & Solution:• Since T = √(2H / g) ∝ 1 / √g and gY = gX / 4, we have TY = 2 TX.
• Given TY = TX + 2 s ⇒ 2 TX = TX + 2 ⇒ TX = 2 s.
• Since v = √(2gH) ∝ √g, we have vY = vX / 2.
• Given vY = vX − 10 ⇒ vX / 2 = vX − 10 ⇒ vX = 20 m s−1.
• Therefore, gX = vX / TX = 20 / 2 = 10 m s−2, and H = (1 / 2) gX TX2 = 20 m.
x-t Graph Parabolic Equation from Vertex & Intercept
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. The position-time (x-t) graph of a uniformly accelerated particle is a parabola that starts at x = +9 m at t = 0, touches the horizontal time axis tangentially (vertex) at t = 3 s (where x = 0), and rises back to x = +9 m at t = 6 s, as shown below. What are the initial velocity u (at t = 0) and the constant acceleration a of the particle?
✔ Vertex Form x(t) = (t − 3)2 = t2 − 6t + 9 Mastered! +4 Marks
✖ Since the vertex is at (t = 3, x = 0) and x(0) = 9, the parabola is x(t) = (t − 3)2 = t2 − 6t + 9, giving u = −6 m s−1 and a = +2 m s−2!
Concept & Solution:• A parabola with its vertex at (t = 3 s, x = 0 m) has the form x(t) = k (t − 3)2.
• Since x(0) = 9 m: 9 = k (0 − 3)2 = 9 k ⇒ k = 1 m s−2.
• Expanding: x(t) = (t − 3)2 = t2 − 6 t + 9.
• Comparing with x(t) = x0 + u t + (1/2) a t2 (or differentiating twice): u = −6 m s−1 and a = +2 m s−2.
Variable Acceleration (a = −k v1/2 Stopping Time vs. Distance Ratio)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. A particle moving with initial speed v0 experiences a resistive deceleration given by a = −k v1/2, where k is a positive constant. If the initial speed v0 is quadrupled (increased to 4 v0), by what factors do its stopping time (Tstop) and its stopping distance (Sstop) increase respectively?
✔ Scaling Laws Tstop ∝ v01/2 & Sstop ∝ v03/2 Mastered! +4 Marks
✖ From dv/dt = −kv1/2, Tstop ∝ v01/2 (increases by 41/2 = 2); from v(dv/dx) = −kv1/2, Sstop ∝ v03/2 (increases by 43/2 = 8)!
Concept & Solution:• Stopping Time: v−1/2 dv = −k dt ⇒ Tstop = (2 / k) v01/2 ∝ v01/2.
• When v0 → 4 v0: Tstop increases by (4)1/2 = 2 times.
• Stopping Distance: v(dv/dx) = −k v1/2 ⇒ v1/2 dv = −k dx ⇒ Sstop = [ 2 / (3k) ] v03/2 ∝ v03/2.
• When v0 → 4 v0: Sstop increases by (4)3/2 = 8 times.
Motion Under Gravity (Two Timings Difference at Height h)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. A ball projected vertically upwards from the ground with initial speed u = 25 m s−1 passes a horizontal wire at height h = 20 m twice (once while going up at t1 and once while coming down at t2). Taking g = 10 m s−2, what is the time interval (Δt = t2 − t1) for which the ball remains above the wire?
✔ Roots of 5t2 − 25t + 20 = 0 (t1 = 1 s, t2 = 4 s ⇒ Δt = 3 s) Mastered! +4 Marks
✖ Solve 20 = 25t − 5t2 ⇒ t2 − 5t + 4 = 0 ⇒ (t − 1)(t − 4) = 0, so the ball is above the wire from t1 = 1 s to t2 = 4 s (Δt = 3 s)!
Concept & Solution:• Setting y(t) = 20 m: 20 = 25 t − (1 / 2)(10) t2.
• Dividing by 5 and rearranging: t2 − 5 t + 4 = 0 ⇒ (t − 1)(t − 4) = 0.
• Thus the ball crosses the wire going up at t1 = 1 s and coming down at t2 = 4 s.
• Time spent above the wire: Δt = t2 − t1 = 4 s − 1 s = 3 s.
Equations of Motion (Ratio of Distances in Equal Time Intervals)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A body starts from rest, accelerates uniformly at rate a for time T (covering distance S1), then moves with the acquired constant velocity for the next time interval T (covering distance S2), and finally decelerates uniformly at the same rate a to come to rest in another time interval T (covering distance S3). What is the ratio (S1 : S2 : S3)?
✔ Triangle-Rectangle-Triangle v-t Area Ratio (1 : 2 : 1) Mastered! +4 Marks
✖ In time T, triangle area is S1 = (1/2) vm T, rectangle area is S2 = vm T = 2 S1, and deceleration triangle area is S3 = (1/2) vm T = S1!
Concept & Solution:• Let peak speed at the end of the first interval be vm = a T.
• 1st interval (triangle): S1 = (1 / 2) vm T.
• 2nd interval (rectangle at constant speed vm): S2 = vm T = 2 S1.
• 3rd interval (symmetric triangle): S3 = (1 / 2) vm T = S1.
• Ratio S1 : S2 : S3 = 1 : 2 : 1.
Variable Acceleration (v2 = a x3 Position-Acceleration Law)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A particle moves along the positive x-axis such that the square of its instantaneous velocity varies as the cube of its position: v2 = 4 x3 (in SI units). What is the instantaneous acceleration of the particle when it passes through x = 2 m?
✔ Implicit Differentiation 2v(dv/dx) = 2a = 12x2 Nailed! +4 Marks
✖ Differentiate v2 = 4x3 with respect to x: 2v(dv/dx) = 12x2 ⇒ a = 6x2 = 6(22) = 24 m s−2!
Concept & Solution:• Differentiating v2 = 4 x3 with respect to x:
• 2 v (dv/dx) = 4 × (3 x2) = 12 x2.
• Since a = v (dv/dx), we immediately obtain: a(x) = 6 x2.
• At x = 2 m: a(2) = 6 × (2)2 = 24 m s−2.
Motion Under Gravity (Ball Thrown Up from Moving Balloon vs. Ground)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. A stone is projected vertically upwards from the top of a tower with an initial speed of u = 20 m s−1 and hits the ground with a speed of v = 40 m s−1. Taking g = 10 m s−2, what are the height (H) of the tower and the total distance (D) travelled by the stone during its entire flight?
✔ Tower Height H = (v2 − u2)/(2g) & Total Distance D = H + 2 h0 Mastered! +4 Marks
✖ H = (402 − 202)/20 = 60 m, and the stone rises h0 = 202/20 = 20 m above the tower before falling 80 m, so Total Distance D = 20 + 80 = 100 m!
Concept & Solution:• Using v2 − u2 = 2 g H between the top of the tower and the ground:
• H = (402 − 202) / (2 × 10) = (1600 − 400) / 20 = 60 m.
• Height risen above the tower before turning around: h0 = u2 / (2 g) = 400 / 20 = 20 m.
• Total Path Distance: D = h0 (up) + (h0 + H) (down) = 20 + 80 = 100 m.
v-t Graph to Speed-Time (|v|-t) Graph Transformation
JEE Main 2022 (June Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A ball is thrown vertically upwards from the ground at t = 0 with speed u = 20 m s−1, reaches its highest point at t = 2 s, and returns to the ground at t = 4 s (taking g = 10 m s−2). Its speed-time (|v|-t) graph is plotted below. What is the slope of the speed-time graph for 0 < t < 2 s and for 2 s < t < 4 s respectively?
✔ Speed Rate d|v|/dt vs. Velocity Rate dv/dt Distinction Mastered! +4 Marks
✖ On a SPEED-time (|v|-t) graph, negative velocities are folded above the time axis into a V-shape: slope d|v|/dt is −10 m s−2 during ascent and +10 m s−2 during descent!
Concept & Solution:• While the velocity-time (v-t) graph is a single straight line of constant negative slope (−10 m s−2), the speed-time (|v|-t) graph is a V-shaped graph above the time axis.
• From 0 to 2 s (slowing down): d|v|/dt = −10 m s−2.
• From 2 to 4 s (speeding up downwards): d|v|/dt = +10 m s−2.
Variable Acceleration (a = 2 t − 1 Minimum Velocity)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. A particle moves along the x-axis with an initial velocity of v(0) = 5 m s−1 at t = 0. Its instantaneous acceleration varies with time t according to a(t) = (4 t − 8) m s−2. What is the minimum velocity attained by the particle, and does the particle ever reverse its direction of motion?
✔ Quadratic Velocity Roots v(t) = 2(t − 1)(t − 3) Solved! +4 Marks
✖ Integrate a(t) = 4t − 8 with v(0) = 5: v(t) = 2t2 − 8t + 5? Wait: 2(2)2 − 8(2) + 5 = −3 m s−1, and roots of 2t2 − 8t + 5 = 0 are two positive times!
Concept & Solution:• Integrating acceleration with v(0) = 5 m s−1:
• v(t) = 5 + ∫0t (4 t − 8) dt = 2 t2 − 8 t + 5.
• Minimum velocity occurs where a(t) = 4t − 8 = 0 ⇒ t = 2 s, giving vmin = 2(4) − 16 + 5 = −3 m s−1.
• Since v(0) = +5 m s−1, v(2) = −3 m s−1, and v(4) = +5 m s−1, velocity crosses zero twice (reversing direction twice).
Set 28 Grand Finale (420-PYQ Milestone: Ball Passing Two Points Symmetry)
JEE Main 2026 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. A ball is projected vertically upwards from the ground. It passes a lower mark A on a vertical pole and takes a total time TA = 6 s (after passing A on the way up) to pass mark A again on the way down. It also passes an upper mark B on the same pole and takes a total time TB = 4 s (after passing B on the way up) to pass mark B again on the way down. Taking g = 10 m s−2, what is the vertical distance (AB) between the two marks A and B?
✔ 420th Question Milestone Conquered! Apex Distance Formula Δh = (g/8)(TA2 − TB2)! +4 Marks
✖ Distance of mark A below the apex is (1/8)g TA2 = 45 m, and distance of mark B below the apex is (1/8)g TB2 = 20 m, so AB = 45 − 20 = 25 m!
Concept & Solution:• If a ball spends total time TA = 6 s above mark A, its time from mark A to the highest point (apex) is tA,up = TA / 2 = 3 s.
• Thus mark A lies at a distance dA = (1 / 2) g (TA / 2)2 = (1 / 8) g TA2 = 5 × 9 = 45 m below the apex.
• Similarly, mark B (TB = 4 s ⇒ tB,up = 2 s) lies at a distance dB = (1 / 8) g TB2 = 5 × 4 = 20 m below the apex.
• Vertical distance between A and B: AB = dA − dB = (g / 8) (TA2 − TB2) = 45 m − 20 m = 25 m.
Set 28 — 420-PYQ Series Completed!
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