Equations of Motion, Motion Under Gravity, Variable Acceleration, and v-t, x-t, a-t Graphs
IIT-JEE MAINS PYQ Arena — Set 22
Equations of Motion, Gravity, Variable Acceleration & Kinematic Graphs
Set 22 • 330-PYQ Series
Equations of Motion (Midpoint Velocity Formula)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. A train accelerating uniformly along a straight track passes a signal post A with speed u = 10 m s−1 and passes the end of the platform B with speed v = 30 m s−1. What is the instantaneous speed of the train when it passes the exact spatial midpoint M between A and B?
✔ Root-Mean-Square Spatial Midpoint Speed Nailed! +4 Marks
✖ 20 m s−1 is the speed at the TIME midpoint! At the SPATIAL midpoint, vM = √((u2 + v2)/2)!
Concept & Solution:• Let total distance AB be 2s. For the first half AM: vM2 − u2 = 2 a s.
• For the second half MB: v2 − vM2 = 2 a s.
• Equating the two expressions gives: vM = √[ (u2 + v2) / 2 ] = √[ (102 + 302) / 2 ] = √500 = 10√5 m s−1.
Question 1 of 15
Motion Under Gravity (Two Timings for Same Height h)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A ball is projected vertically upwards from the ground. It crosses a point P at a height h above the ground at time t1 = 2 s while ascending, and crosses the exact same point P again at time t2 = 6 s (measured from the instant of projection) while descending. Taking g = 10 m s−2, what are the height h of point P and the maximum height Hmax reached by the ball?
✔ Vieta’s Roots Formula h = (1/2) g t1 t2 Mastered! +4 Marks
✖ From (1/2)gt2 − ut + h = 0, product of roots t1t2 = 2h/g ⇒ h = (1/2)(10)(2)(6) = 60 m, and T = t1 + t2 = 8 s ⇒ Hmax = gT2/8 = 80 m!
Concept & Solution:• The quadratic equation for height h is: (1/2) g t2 − u t + h = 0, whose roots are t1 = 2 s and t2 = 6 s.
• Product of roots: t1 t2 = 2 h / g ⇒ h = (1/2) g t1 t2 = 0.5 × 10 × 2 × 6 = 60 m.
• Sum of roots (total time of flight): T = t1 + t2 = 2 + 6 = 8 s (so u = g T / 2 = 40 m s−1).
• Maximum Height: Hmax = (1/8) g T2 = (10 × 64) / 8 = 80 m.
v-t Graph Analysis (Distance vs. Displacement)
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 a straight line is shown below. From t = 0 to t = 4 s, the graph forms a positive triangle of peak height +20 m s−1, and from t = 4 s to t = 6 s, it forms a negative triangle of peak depth −10 m s−1. What is the ratio of the Total Distance travelled to the Net Displacement over the 6 s interval?
✔ Signed Area vs. Absolute Area Ratio (50 m / 30 m = 5 : 3) Nailed! +4 Marks
✖ Positive area A1 = 40 m and negative area magnitude |A2| = 10 m, so Distance = 40 + 10 = 50 m and Displacement = 40 − 10 = 30 m!
Concept & Solution:• Area of positive triangle (0 to 4 s): A1 = (1/2) × 4 s × 20 m s−1 = 40 m.
• Magnitude of negative triangle (4 to 6 s): |A2| = (1/2) × 2 s × 10 m s−1 = 10 m.
• Total Distance = A1 + |A2| = 40 + 10 = 50 m.
• Net Displacement = A1 − |A2| = 40 − 10 = 30 m ⇒ Ratio = 50 / 30 = 5 : 3.
Equations of Motion (α-β Acceleration-Deceleration Problem)
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. A metro train starts from rest at Station P, accelerates uniformly at rate α = 2 m s−2 for some time, and then immediately decelerates uniformly at rate β = 4 m s−2 to come to rest at Station Q. If the total time elapsed between P and Q is T = 12 s, what are the maximum speed (vmax) reached and the total distance (S) between the two stations?
✔ Classic αβ/(α+β) Triangle v-t Formula Mastered! +4 Marks
✖ Effective acceleration aeff = αβ/(α+β) = 8/6 = 4/3 m s−2, so vmax = aeff T = 16 m s−1 and S = (1/2) vmax T = 96 m!
Concept & Solution:• Since vmax = α t1 = β t2 and t1 + t2 = T:
• vmax = [ α β / (α + β) ] × T = [ (2 × 4) / (2 + 4) ] × 12 = (8 / 6) × 12 = 16 m s−1.
• Total Distance = Area of v-t triangle = (1 / 2) × vmax × T = 0.5 × 16 × 12 = 96 m.
Motion Under Gravity (Last Second Distance = First 3 Seconds Distance)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. A stone is dropped from rest from the top of a tall cliff. If the distance covered by the stone during the last second of its fall is equal to the distance covered by it during the first 3 seconds of its fall, what is the total height H of the cliff? (Take g = 10 m s−2)
✔ Galileo’s Odd-Integer Equating (2n − 1 = 32) Nailed! +4 Marks
✖ Equate Sn = (g/2)(2n − 1) to S3 = (g/2)(32) ⇒ 2n − 1 = 9 ⇒ n = 5 s, giving H = (1/2)(10)(52) = 125 m!
Concept & Solution:• Let the total time of fall be n seconds.
• Distance in the first 3 seconds: S3 = (1/2) g (3)2 = 9 (g / 2).
• Distance in the nth (last) second: Sn = (g / 2) (2n − 1).
• Equating Sn = S3 ⇒ 2n − 1 = 9 ⇒ n = 5 s.
• Total Height of cliff: H = (1/2) g n2 = 0.5 × 10 × 25 = 125 m.
v2 vs. x Graph Slope & Acceleration
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. For a particle moving along the x-axis, the graph of the square of its velocity (v2 in m2 s−2) versus position (x in m) is a straight line cutting the vertical axis at v2 = 64 m2 s−2 (at x = 0) and cutting the horizontal axis at x = 16 m (where v2 = 0). What is the magnitude of the acceleration of the particle, and how much time does it take to come to rest at x = 16 m?
✔ Half-Slope Rule a = (1/2) × d(v2)/dx Mastered! +4 Marks
✖ Since v2 = u2 + 2ax, acceleration is HALF the slope of the v2-x graph: a = (1/2)(−64/16) = −2 m s−2!
Concept & Solution:• Equation of the straight-line graph: v2 = 64 − (64 / 16) x = 64 − 4 x.
• Comparing with v2 = u2 + 2 a x gives: u = √64 = 8 m s−1 and 2 a = −4 ⇒ a = −2 m s−2 (|a| = 2 m s−2).
• Time to come to rest: t = u / |a| = 8 m s−1 / 2 m s−2 = 4 s.
Variable Acceleration (a-t Graph Area & Final Speed)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A particle starts with an initial velocity of u = 5 m s−1 at t = 0. Its acceleration-time (a-t) graph consists of a constant acceleration of 4 m s−2 from t = 0 to t = 4 s, followed by a linear decrease to 0 m s−2 at t = 8 s. What is the instantaneous velocity of the particle at t = 8 s?
✔ Area Under a-t Graph = Δv + Initial Velocity u Mastered! +4 Marks
✖ Don’t forget to add the initial velocity u = 5 m s−1 to the trapezoidal area under the a-t graph (24 m s−1)!
Concept & Solution:• Change in velocity Δv = ∫08 a(t) dt = Area under a-t graph.
• Area of rectangle (0 to 4 s) + Area of triangle (4 to 8 s) = (4 × 4) + (1/2 × 4 × 4) = 16 + 8 = 24 m s−1.
• Final Velocity at t = 8 s: v(8) = u + Δv = 5 + 24 = 29 m s−1.
Equations of Motion (Braking Stopping Distance & Reaction Time)
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. When an automobile moves at a speed of 10 m s−1, the total stopping distance (including the driver’s reaction-time distance and the constant-deceleration braking distance) is 15 m. When the same automobile moves at 20 m s−1, the total stopping distance is 50 m. What are the driver’s reaction time (tr) and the constant braking deceleration (a)?
✔ Linear Reaction Distance + Quadratic Braking Distance Solved! +4 Marks
✖ Use S = u tr + u2/(2a): 10 tr + 50/a = 15 and 20 tr + 200/a = 50!
Concept & Solution:• Total stopping distance formula: S = u tr + u2 / (2 a).
• For u1 = 10 m s−1: 10 tr + 100 / (2 a) = 15 ⇒ 20 tr + 100 / a = 30.
• For u2 = 20 m s−1: 20 tr + 400 / (2 a) = 50 ⇒ 20 tr + 200 / a = 50.
• Subtracting the two equations gives: 100 / a = 20 ⇒ a = 5 m s−2, and 10 tr + 10 = 15 ⇒ tr = 0.5 s.
Motion Under Gravity (Three-Throw Tower Height Formula)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. From the top of a tower of height H, a ball thrown vertically upwards with speed u takes time t1 = 6 s to reach the ground. A second ball thrown vertically downwards from the same point with the same speed u takes time t2 = 2 s to reach the ground. How much time (t3) will a third ball take to reach the ground if it is simply dropped from rest from the top of the tower, and what is the height H of the tower? (Take g = 10 m s−2)
✔ Geometric Mean Time t3 = √(t1 t2) Mastered! +4 Marks
✖ For upward, downward, and free-fall drops from height H, t3 = √(t1 t2) = √(6 × 2) = 2√3 s, and H = (1/2) g t1 t2 = 60 m!
Concept & Solution:• Upward throw: H = −u t1 + (1/2) g t12.
• Downward throw: H = +u t2 + (1/2) g t22.
• Eliminating u gives: H = (1 / 2) g t1 t2 = 0.5 × 10 × 6 × 2 = 60 m.
• Since H = (1 / 2) g t32 for a ball dropped from rest, t3 = √(t1 t2) = √12 = 2√3 s.
x-t Graph Concavity & Sign of Velocity/Acceleration
JEE Main 2022 (June Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. The position-time (x-t) graph of a particle moving in a straight line is a smooth curve that is rising (positive slope) but concave downwards (bending towards the time axis) throughout the interval 0 < t < T. Which of the following correctly describes the signs of its instantaneous velocity (v) and instantaneous acceleration (a) in this interval?
✔ First Derivative (Slope = v) & Second Derivative (Concavity = a) Mastered! +4 Marks
✖ Rising x-t graph means dx/dt = v > 0, while concave-downward curvature means d2x/dt2 = a < 0!
Concept & Solution:• Instantaneous velocity is the slope of the x-t graph: since the curve is rising, v = dx/dt > 0.
• Instantaneous acceleration is the rate of change of slope (concavity): since the curve is concave downwards (slope is decreasing), a = d2x/dt2 < 0.
Variable Acceleration (v-x Linear Graph Slope)
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. The velocity-position (v-x) graph of a particle moving along the x-axis is a straight line passing through (x = 0, v = 20 m s−1) and (x = 40 m, v = 0 m s−1). What is the instantaneous acceleration of the particle when it passes through the midpoint x = 20 m?
✔ Value × Slope Rule a = v (dv/dx) Mastered! +4 Marks
✖ On a v-x graph, acceleration is NOT just the slope! It is a = v × (dv/dx) = (10 m s−1) × (−0.5 s−1) = −5.0 m s−2!
Concept & Solution:• Equation of the v-x line: v(x) = 20 − (20 / 40) x = 20 − 0.5 x.
• Slope of the v-x graph: dv/dx = −0.5 s−1.
• At x = 20 m, instantaneous velocity is v(20) = 20 − 0.5(20) = 10 m s−1.
• Instantaneous acceleration: a = v (dv/dx) = 10 × (−0.5) = −5.0 m s−2.
Motion Under Gravity (Juggler’s Balls in Air)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. A juggler throws balls vertically upwards at regular time intervals, throwing the next ball at the exact instant the previous ball reaches the highest point (apex) of its path. If he throws n = 2 balls per second, what is the maximum height (H) reached by each ball above his hands? (Take g = 10 m s−2)
✔ Juggler Apex-Interval Formula H = g / (2 n2) Mastered! +4 Marks
✖ Throwing 2 balls per second means the time to reach the apex is Δt = 1/2 = 0.5 s, so H = (1/2) g (Δt)2 = 5 × 0.25 = 1.25 m!
Concept & Solution:• Since n = 2 balls are thrown per second, the time interval between consecutive throws is Δt = 1 / n = 0.5 s.
• Because each ball reaches its apex when the next ball is thrown, the time of ascent is ta = u / g = Δt = 0.5 s.
• Maximum height: H = (1 / 2) g ta2 = g / (2 n2) = 10 / (2 × 4) = 1.25 m.
Equations of Motion (Bullet Penetration in Wooden Planks)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A bullet fired into a fixed target loses (1 / 20)th of its initial velocity after penetrating a single wooden plank of thickness d. Assuming uniform retardation offered by the planks, what is the minimum number of such identical planks required to completely stop the bullet?
✔ Plank Stopping Formula N = n2 / (2n − 1) Mastered! +4 Marks
✖ For losing 1/n of velocity in 1 plank, number of planks is n2/(2n − 1) = 400 / 39 = 10.25 ⇒ rounded UP to 11 planks!
Concept & Solution:• After 1 plank of thickness d, speed becomes v1 = u − u / 20 = (19 / 20) u.
• Using v2 − u2 = −2 a s for 1 plank: u2 − (19/20)2 u2 = (39 / 400) u2 = 2 a d.
• For N planks to stop the bullet (v = 0): u2 = 2 a (N d).
• Dividing the two equations gives: N = 400 / 39 ≈ 10.26 ⇒ minimum integer planks required = 11 planks.
Motion Under Gravity (Parachutist Two-Stage Descent)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. A parachutist bails out from an aircraft and falls freely under gravity (g = 10 m s−2) for a vertical distance of 50 m before opening his parachute. Once the parachute opens, he decelerates uniformly at 2 m s−2 and reaches the ground with a safe speed of 2 m s−1. At what total height H above the ground did he bail out from the aircraft?
✔ Free-Fall + Deceleration Height Sum (50 m + 249 m = 299 m) Nailed! +4 Marks
✖ Speed after 50 m free fall is v12 = 2(10)(50) = 1000. Deceleration distance h2 = (1000 − 4) / (2 × 2) = 249 m, plus the initial 50 m = 299 m!
Concept & Solution:• Stage 1 (Free fall for h1 = 50 m): v12 = 2 g h1 = 2 × 10 × 50 = 1000 m2 s−2.
• Stage 2 (Uniform deceleration a = −2 m s−2 to final speed v2 = 2 m s−1):
• h2 = (v12 − v22) / (2 |a|) = (1000 − 22) / (2 × 2) = 996 / 4 = 249 m.
• Total Bail-Out Height: H = h1 + h2 = 50 m + 249 m = 299 m.
Set 22 Grand Finale (a-x Graph Area & Final Speed)
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. A particle starts from rest at x = 0 and moves along the positive x-axis. Its acceleration-position (a-x) graph is a triangle whose base extends from x = 0 to x = 10 m and whose peak height is amax = 5 m s−2 (at x = 5 m). What is the instantaneous speed of the particle when it reaches x = 10 m?
✔ 330th Question Milestone Conquered! Area Under a-x Graph = (v2 − u2)/2! +4 Marks
✖ Since v dv = a dx, the area under the a-x graph equals (v2 − u2)/2 = 25, so v2 = 50 ⇒ v = 5√2 m s−1!
Concept & Solution:• From v dv = a(x) dx, integrating both sides gives:
• (v2 − u2) / 2 = ∫010 a(x) dx = Area under the a-x graph.
• Area of the triangle = (1 / 2) × Base × Height = (1 / 2) × 10 m × 5 m s−2 = 25 m2 s−2.
• Since u = 0: v2 / 2 = 25 ⇒ v2 = 50 ⇒ v = √50 = 5√2 m s−1.
Set 22 — 330-PYQ Series Completed!
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