Equations of Motion, Motion Under Gravity, Variable Acceleration, and v-t, x-t, a-t Graphs
IIT-JEE MAINS PYQ Arena — Set 27
Equations of Motion, Gravity, Variable Acceleration & Kinematic Graphs
Set 27 • 405-PYQ Milestone
Equations of Motion (Consecutive Equal Distances with Different Times)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. A particle moving with constant acceleration along a straight line covers a distance of 12 m in the first 2 seconds (from t = 0 to t = 2 s) and another 12 m in the next 4 seconds (from t = 2 s to t = 6 s). What are the initial velocity (u) and the constant acceleration (a) of the particle?
✔ Cumulative Displacement Equations S(2) = 12 m & S(6) = 24 m Solved! +4 Marks
✖ Always write cumulative displacements from t = 0! S(2) = 2u + 2a = 12 and S(6) = 6u + 18a = 24!
Concept & Solution:• From t = 0 to t = 2 s (S1 = 12 m): 12 = u(2) + (1/2) a (2)2 ⇒ u + a = 6.
• From t = 0 to t = 6 s (Stotal = 12 + 12 = 24 m): 24 = u(6) + (1/2) a (6)2 ⇒ u + 3 a = 4.
• Subtracting the two equations gives: 2 a = −2 ⇒ a = −1.0 m s−2 and u = 7 m s−1.
Question 1 of 15
Motion Under Gravity (Last-Second Distance of Upward Projection)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. Two balls A and B of masses 1 kg and 2 kg are projected vertically upwards from the ground with initial speeds of uA = 35 m s−1 and uB = 70 m s−1 respectively. Neglecting air resistance and taking g = 10 m s−2, what are the distances covered by Ball A and Ball B during the last second of their upward ascent (just before reaching their respective highest points)?
✔ Universal Last-Second Ascent Invariance (g / 2 = 5.0 m) Mastered! +4 Marks
✖ By time-reversal symmetry around the apex (where v = 0), the distance covered in the last 1 s of ascent equals the distance covered in the first 1 s of free fall from rest: (1/2)g(1)2 = 5.0 m for ANY initial speed u!
Concept & Solution:• At the highest point (apex), instantaneous velocity is v = 0.
• By time-reversal symmetry, the distance travelled during the last τ seconds of upward ascent before stopping at the apex is identical to the distance fallen from rest during the first τ seconds of descent:
• Slast τ = (1 / 2) g τ2.
• For τ = 1 s and g = 10 m s−2: S = 0.5 × 10 × (1)2 = 5.0 m for both balls, completely independent of u and m!
v-t Graph Trapezoidal Area with Negative Lobe
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. The velocity-time (v-t) graph of a particle moving along the x-axis is shown below. From t = 0 to t = 5 s, the graph is a positive trapezium with parallel bases of 5 s and 2 s and height +10 m s−1. From t = 5 s to t = 7 s, it is a negative rectangle of constant velocity −10 m s−1. What are the Total Distance and Net Displacement over the 7 s interval?
✔ Trapezium Area (35 m) ± Rectangle Area (20 m) Mastered! +4 Marks
✖ Positive trapezium area A1 = 0.5(5 + 2)(10) = 35 m and negative rectangle magnitude |A2| = 2 × 10 = 20 m!
Concept & Solution:• Area of positive trapezium (0 to 5 s): A1 = (1 / 2) × (5 + 2) s × 10 m s−1 = 35 m.
• Magnitude of negative rectangle (5 to 7 s): |A2| = (7 − 5) s × 10 m s−1 = 20 m.
• Total Distance = A1 + |A2| = 35 + 20 = 55 m.
• Net Displacement = A1 − |A2| = 35 − 20 = 15 m.
Variable Acceleration (v(t) = k t e−α t Zero Acceleration Time)
JEE Main 2025 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A particle moves along a straight line such that its instantaneous velocity at time t is given by v(t) = A t e−t / τ, where A and τ are positive constants. At what instant of time t does its instantaneous acceleration become zero, and what is the maximum velocity (vmax) attained by the particle?
✔ Product Rule Differentiation a(t) = A e−t/τ(1 − t/τ) = 0 Mastered! +4 Marks
✖ Differentiate using the product rule: dv/dt = A e−t/τ (1 − t/τ) = 0 ⇒ t = τ, where v(τ) = A τ e−1 = Aτ / e!
Concept & Solution:• Using the product rule to find instantaneous acceleration:
• a(t) = dv/dt = A [ 1 · e−t / τ + t · (−1 / τ) e−t / τ ] = A e−t / τ (1 − t / τ).
• Setting a(t) = 0 gives: 1 − t / τ = 0 ⇒ t = τ.
• Maximum velocity at t = τ: vmax = v(τ) = A τ e−1 = (A τ) / e.
Motion Under Gravity (Balloon Ascending with Acceleration)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. A hot-air balloon starts rising from rest from the ground with a constant upward acceleration of a = 1.25 m s−2. At the end of t1 = 8 s, a stone is gently released from the balloon. Taking g = 10 m s−2, how much time (t2) after its release will the stone take to reach the ground?
✔ Release Height (40 m) & Upward Speed (+10 m s−1) Tracked Perfectly! +4 Marks
✖ At t1 = 8 s, balloon height is h = 0.5(1.25)(64) = 40 m and upward speed is u = 1.25(8) = +10 m s−1. Solving −40 = 10t2 − 5t22 gives t2 = 4 s!
Concept & Solution:• At t1 = 8 s (instant of release):
• Height of balloon: h = (1 / 2) a t12 = 0.5 × 1.25 × (8)2 = 40 m.
• Upward velocity inherited by stone: u = a t1 = 1.25 × 8 = +10 m s−1.
• For the stone to reach the ground (Δy = −40 m):
• −40 = 10 t2 − (1 / 2)(10) t22 ⇒ t22 − 2 t2 − 8 = 0 ⇒ (t2 − 4)(t2 + 2) = 0 ⇒ t2 = 4 s.
Equations of Motion (Braking Distance after Losing Half Speed)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. A bullet fired into a thick wooden target with initial velocity u loses half of its velocity (so its speed becomes u / 2) after penetrating a distance of 30 cm. Assuming constant retardation inside the target, how much additional distance (Δx) will the bullet penetrate before coming to complete rest?
✔ Kinetic Energy Loss 3/4 vs. 1/4 Ratio (30 cm / 3 = 10 cm) Mastered! +4 Marks
✖ In losing half its speed (u → u/2), u2 − u2/4 = (3/4)u2 is lost in 30 cm! The remaining (1/4)u2 takes 1/3rd of 30 cm = 10 cm!
Concept & Solution:• For the first 30 cm (speed drops from u to u / 2):
• u2 − (u / 2)2 = (3 / 4) u2 = 2 a (30 cm).
• For the remaining distance Δx (speed drops from u / 2 to 0):
• (u / 2)2 − 0 = (1 / 4) u2 = 2 a (Δx).
• Dividing the two equations gives: Δx / 30 cm = (1 / 4) / (3 / 4) = 1 / 3 ⇒ Δx = 10 cm.
a-x Graph with Negative Deceleration Lobe
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A particle starts from rest at x = 0 and moves along the positive x-axis. Its acceleration-position (a-x) graph is a constant +4 m s−2 from x = 0 to x = 8 m, and then switches to a constant −2 m s−2 for x > 8 m, as shown below. What is the maximum speed attained by the particle, and at what position xstop does the particle come to rest again?
✔ Zero Net Area Under a-x Graph for Rest-to-Rest Motion Mastered! +4 Marks
✖ Since (v2 − 0)/2 = ∫ a dx, maximum speed occurs at x = 8 m (vmax2/2 = 32 ⇒ vmax = 8 m s−1), and it stops when negative area equals +32 (at x = 8 + 16 = 24 m)!
Concept & Solution:• Since v2 / 2 = ∫0x a(x) dx, speed is maximum at x = 8 m where acceleration changes from + to −:
• vmax2 / 2 = 4 × 8 = 32 ⇒ vmax2 = 64 ⇒ vmax = 8 m s−1.
• The particle comes to rest again (v = 0) when the net signed area under the a-x graph becomes zero:
• 32 − 2 (xstop − 8) = 0 ⇒ xstop − 8 = 16 ⇒ xstop = 24 m.
Variable Acceleration (v = v0(1 − x/x0) Exponential Time)
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. A particle starts from the origin (x = 0) at t = 0 and moves along the positive x-axis such that its velocity decreases linearly with position x according to v(x) = v0 (1 − x / L), where v0 and L are positive constants. What is the position x(t) of the particle as a function of time t, and how much time does it take to reach x = L / 2?
✔ Linear v-x Graph → Exponential x(t) Approach Mastered! +4 Marks
✖ Separate variables: dx / (1 − x/L) = v0 dt ⇒ −L ln(1 − x/L) = v0 t ⇒ x(t) = L(1 − e−v0t/L)!
Concept & Solution:• Since dx/dt = v0 (1 − x / L), separating variables gives:
• dx / (1 − x / L) = v0 dt.
• Integrating from (t = 0, x = 0) to (t, x): −L ln(1 − x / L) = v0 t ⇒ x(t) = L (1 − e−v0 t / L).
• For x = L / 2: e−v0 t / L = 1 / 2 ⇒ t1/2 = (L / v0) ln(2) (and it takes infinite time to reach x = L!).
Motion Under Gravity (Two Stones Thrown Up & Down Simultaneously)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. From the top of a high cliff, Stone 1 is thrown vertically upwards with speed u = 15 m s−1 and simultaneously Stone 2 is thrown vertically downwards with speed v = 10 m s−1. Before either stone hits the ground, how does the vertical distance Δy(t) between the two stones vary with time t, and what is their separation at t = 3 s?
✔ Zero Relative Acceleration (g Cancels Out) Mastered! +4 Marks
✖ Since both stones are in free flight under gravity, (1/2)gt2 cancels out completely! Relative separation grows linearly at (15 + 10)t = 25t!
Concept & Solution:• Since both stones experience the exact same downward acceleration g, their relative acceleration is arel = 0.
• Relative speed of separation: vrel = u + v = 15 + 10 = 25 m s−1.
• Separation grows linearly with time: Δy(t) = vrel t = 25 t.
• At t = 3 s: Δy(3) = 25 × 3 = 75 m.
Equations of Motion (Maximum Separation Between Two Moving Cars)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. At t = 0, Car A passes a checkpoint moving at a constant speed of vA = 24 m s−1. At the exact same instant, Car B starts from rest from a point 10 m behind the checkpoint and accelerates in the same direction at a constant rate of aB = 6 m s−2. What is the maximum separation (Dmax) between Car A and Car B before Car B overtakes Car A?
✔ Equal-Velocity Condition (vB = vA at t = 4 s) + Initial 10 m Lead Mastered! +4 Marks
✖ Maximum separation occurs when vB = vA (6t = 24 ⇒ t = 4 s)! Additional lead gained = 24(4) − 0.5(6)(16) = 48 m, plus initial 10 m = 58 m!
Concept & Solution:• Separation between the two cars is D(t) = D0 + vA t − (1 / 2) aB t2.
• Maximum separation occurs when dD/dt = 0 ⇒ vB(t) = aB t = vA:
• 6 t = 24 ⇒ t = 4 s.
• Maximum Separation: Dmax = 10 + 24(4) − 0.5(6)(4)2 = 10 + 96 − 48 = 58 m.
Motion Under Gravity (Free-Fall Distance in Last Second = First 2 Seconds)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. A ball is dropped from rest from the top of a tower. The distance covered by the ball in the last second of its motion is equal to the distance covered by it in the first 2 seconds of its motion. Taking g = 10 m s−2, what are the total time of fall (n) and the height (H) of the tower?
✔ Non-Integer Total Time n = 2.5 s & H = 31.25 m Nailed! +4 Marks
✖ Equate Slast 1s = (g/2)(2n − 1) to S2 = (g/2)(22) ⇒ 2n − 1 = 4 ⇒ n = 2.5 s, so H = 5(2.5)2 = 31.25 m!
Concept & Solution:• Distance covered in the first 2 seconds: S2 = (1 / 2) g (2)2 = 4 (g / 2) = 20 m.
• Distance covered in the last 1 second (from t = n − 1 to t = n): Slast = (g / 2) (2n − 1).
• Equating the two: 2n − 1 = 4 ⇒ n = 2.5 s.
• Total Height of tower: H = (1 / 2) g n2 = 5 × (2.5)2 = 5 × 6.25 = 31.25 m.
Piecewise Parabolic x-t Graph to Velocity & Distance
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. The position-time (x-t) graph of a particle is a downward-opening parabola given by x(t) = (8 t − 2 t2) m for 0 ≤ t ≤ 4 s, as shown below. At what instant of time does the particle reverse its direction of motion, and what are its Net Displacement and Total Distance travelled over the 4 s interval?
✔ Parabolic Vertex on x-t Graph = Turning Point (v = 0) Mastered! +4 Marks
✖ At the vertex t = 2 s, slope dx/dt = 8 − 4t = 0 and x(2) = 8 m! The particle moves 0 → 8 m → 0, so Displacement = 0 m and Distance = 16 m!
Concept & Solution:• Instantaneous velocity: v(t) = dx/dt = 8 − 4 t = 0 at t = 2 s (vertex of the x-t parabola).
• Position at t = 0: x(0) = 0; at t = 2 s: x(2) = 8(2) − 2(4) = 8 m; at t = 4 s: x(4) = 0.
• Net Displacement = x(4) − x(0) = 0 m.
• Total Distance = |8 − 0| + |0 − 8| = 8 + 8 = 16 m.
Variable Acceleration (Logarithmic Displacement from a = −k v2)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A motorboat moving with initial velocity v0 at t = 0 (at x = 0) has its engine cut off and experiences a quadratic water-drag deceleration given by dv/dt = −k v2, where k is a positive constant. What are its instantaneous velocity v(t) and its displacement x(t) as functions of time t?
✔ Two-Step Integration dv/v2 = −k dt ⇒ x(t) = (1/k) ln(1 + k v0 t) Mastered! +4 Marks
✖ First integrate −dv/v2 = k dt to get 1/v − 1/v0 = kt ⇒ v(t) = v0/(1 + kv0t), then integrate v(t) dt to get (1/k) ln(1 + kv0t)!
Concept & Solution:• Step 1: Separating variables in dv/dt = −k v2 gives −dv / v2 = k dt.
• Integrating from v0 to v: 1 / v − 1 / v0 = k t ⇒ v(t) = v0 / (1 + k v0 t).
• Step 2: Integrating dx = v(t) dt from t = 0 to t:
• x(t) = ∫0t [ v0 / (1 + k v0 t) ] dt = (1 / k) ln(1 + k v0 t).
Equations of Motion (Two Particles Meeting After Opposite Launches)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. A ball is projected vertically upwards from the top of a tower of height H = 40 m with an initial speed of u = 10 m s−1. Taking g = 10 m s−2, what is the ratio of the time taken by the ball to return to the level of the top of the tower to the total time taken by it to hit the ground at the base of the tower?
✔ Return-to-Top Time (2 s) vs. Ground Impact Time (4 s) = 1 : 2 Nailed! +4 Marks
✖ Time to return to tower top is ttop = 2u/g = 2 s, and total time to hit ground (−40 = 10t − 5t2) is ttotal = 4 s, giving ratio 1 : 2!
Concept & Solution:• Time to return to the level of projection (top of tower): ttop = 2 u / g = (2 × 10) / 10 = 2 s.
• Total time to reach the ground (Δy = −40 m):
• −40 = 10 t − 5 t2 ⇒ t2 − 2 t − 8 = 0 ⇒ (t − 4)(t + 2) = 0 ⇒ ttotal = 4 s.
• Ratio ttop : ttotal = 2 s : 4 s = 1 : 2.
Set 27 Grand Finale (405-PYQ Milestone: Symmetric v-t Acceleration & Distance)
JEE Main 2026 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. A lift moves vertically upwards from the ground floor (rest) to the top floor (rest) of a skyscraper of height H = 120 m. Its maximum permissible upward acceleration is 2 m s−2, its maximum permissible upward cruising speed is 10 m s−1, and its maximum permissible retardation magnitude is 2 m s−2. What is the minimum possible time (Tmin) in which the lift can complete the 120 m ascent?
✔ 405th Question Milestone Conquered! Optimal Trapezoidal v-t Profile Solved! +4 Marks
✖ Acceleration to 10 m s−1 takes 5 s (25 m) and deceleration takes 5 s (25 m). The remaining 70 m at 10 m s−1 takes 7 s, so Tmin = 5 + 7 + 5 = 17.0 s!
Concept & Solution:• For minimum time, the lift accelerates at the maximum rate (2 m s−2) to reach vmax = 10 m s−1, cruises at 10 m s−1, and decelerates at 2 m s−2:
• Acceleration time: t1 = 10 / 2 = 5.0 s; distance covered: s1 = (1/2)(2)(5)2 = 25 m.
• Deceleration time: t3 = 10 / 2 = 5.0 s; distance covered: s3 = 25 m.
• Remaining cruising distance at 10 m s−1: s2 = 120 − (25 + 25) = 70 m ⇒ t2 = 70 / 10 = 7.0 s.
• Minimum Total Time: Tmin = 5.0 + 7.0 + 5.0 = 17.0 s.
Set 27 — 405-PYQ Milestone Completed!
Phenomenal achievement! You now have 405 unique, zero-dependency JEE Main PYQs across Sets 1–27. Check your Set 27 score below or click any question pill above to review.
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