TOPIC: Frame of Reference, Distance, Displacement, Average & Instantaneous Velocity/Acceleration
IIT-JEE MAINS PYQ Arena — Set 19
Frame of Reference, Distance, Displacement & Avg/Instantaneous Kinematics
Set 19 • Kinematics V
Average Speed with Two Equal Time Splits
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. A bus travels between two towns. During the first half of the total journey time, it moves with speed v1 = 25 km h−1. During the second half of the total journey time, it covers half the remaining distance at v2 = 30 km h−1 and the other half of the remaining distance at v3 = 70 km h−1. What is the average speed of the bus for the entire journey?
✔ Two-Stage Harmonic + Arithmetic Mean Nailed! +4 Marks
✖ First find the harmonic mean of v2 and v3 (equal distances ⇒ 42 km h−1), then take the arithmetic mean with v1 = 25 km h−1!
Concept & Solution:• For the second half of the time (split into two equal distances at 30 km h−1 and 70 km h−1):
• v23,avg = (2 v2 v3) / (v2 + v3) = (2 × 30 × 70) / (30 + 70) = 4200 / 100 = 42 km h−1.
• Now the two halves of the total journey time are covered at 25 km h−1 and 42 km h−1:
• Total Average Speed = (25 + 42) / 2 = 67 / 2 = 33.5 km h−1.
Question 1 of 15
Frame of Reference (Mid-Air Collision of Two Projectiles)
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. From the top of a 100 m high vertical tower, Ball A is dropped from rest at t = 0. At the exact same instant (t = 0), Ball B is projected vertically upwards from the base of the tower with an initial speed of 50 m s−1. Working in the reference frame of Ball A, at what time t and at what height h above the ground do the two balls collide? (Take g = 10 m s−2)
✔ Zero Relative Gravity (grel = 0) Shortcut Mastered! +4 Marks
✖ In Ball A’s frame, relative acceleration is zero, so t = H / urel = 100 / 50 = 2 s, and Ball A falls 20 m (height = 80 m)!
Concept & Solution:• Relative acceleration between the two balls in free flight: aBA = (−g) − (−g) = 0.
• Initial relative velocity of approach: urel = 50 − 0 = 50 m s−1.
• Time of collision: t = H / urel = 100 m / 50 m s−1 = 2 s.
• Distance fallen by Ball A in 2 s: d = (1/2) g t2 = 5 × (2)2 = 20 m ⇒ Height above ground h = 100 − 20 = 80 m.
Instantaneous Acceleration from Reciprocal Velocity 1/v(x)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. For a particle moving along the positive x-axis, the reciprocal of its instantaneous velocity (1 / v) varies linearly with its position x as 1 / v = 3 x + 2 (in SI units). What is the instantaneous acceleration of the particle when it passes through the position x = 2 m?
✔ Reciprocal Derivative d(1/v)/dx = −(1/v2)(dv/dx) Mastered! +4 Marks
✖ Differentiate 1/v = 3x + 2 with respect to x: −(1/v2)(dv/dx) = 3 ⇒ a = v(dv/dx) = −3 v3!
Concept & Solution:• Differentiating v−1 = 3 x + 2 with respect to x:
• −v−2 (dv/dx) = 3 ⇒ dv/dx = −3 v2.
• Therefore, instantaneous acceleration: a = v (dv/dx) = −3 v3.
• At x = 2 m: 1/v = 3(2) + 2 = 8 ⇒ v = 1 / 8 m s−1.
• Acceleration at x = 2 m: a = −3 × (1 / 8)3 = −3 / 512 m s−2.
Frame of Reference (Rain-Woman Umbrella Direction Reversal)
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. When a woman walks due East on a horizontal road at a speed of 3 km h−1, raindrops appear to fall vertically downwards on her. When she increases her speed to 6 km h−1 due East, the raindrops appear to hit her at an angle of 45° with the vertical. What is the actual speed of the raindrops with respect to the ground?
✔ Two-Speed Rain Vector Decomposition Nailed! +4 Marks
✖ At 3 km h−1 East, rain appears vertical, so vrx = +3 km h−1. At 6 km h−1 East, tan 45° = |3 − 6| / vry = 1 ⇒ vry = 3 km h−1!
Concept & Solution:• Let true rain velocity in the ground frame be vr = vrx î − vry ĵ.
• When walking at vw1 = 3 î km h−1, relative velocity vrw1 = (vrx − 3) î − vry ĵ is purely vertical ⇒ vrx = 3 km h−1.
• When running at vw2 = 6 î km h−1, vrw2 = (3 − 6) î − vry ĵ = −3 î − vry ĵ makes 45° with vertical ⇒ tan 45° = 3 / vry ⇒ vry = 3 km h−1.
• True speed of rain: |vr| = √(32 + 32) = 3√2 km h−1.
Distance vs. Displacement in 3D Helical Motion
JEE Main 2023 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. A charged particle moves along a 3D helix such that its position vector at time t is given by r(t) = (3 cos(π t) î + 3 sin(π t) ĵ + 4 π t k̂) m, where t is in seconds. What are the Total Path Distance travelled and the magnitude of the Net Displacement of the particle from t = 0 s to t = 2 s?
✔ 3D Helical Arc Length vs. Axial Pitch Displacement Mastered! +4 Marks
✖ Speed |v| = √((3π)2 + (4π)2) = 5π m s−1, so in 2 s (one full turn) Distance = 10π m and Net Displacement = Δz = 8π m!
Concept & Solution:• Velocity vector: v(t) = dr/dt = −3π sin(π t) î + 3π cos(π t) ĵ + 4π k̂.
• Instantaneous Speed (constant): |v| = √((3π)2 + (4π)2) = 5π m s−1.
• Total Path Distance in Δt = 2 s: d = |v| × Δt = 5π × 2 = 10π m.
• At t = 0: r(0) = 3 î; at t = 2 s (one full turn): r(2) = 3 î + 8π k̂ ⇒ |Δr| = 8π m.
Separation of Variables (a = −k √v) Stopping Distance
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. A boat moving with an initial speed of v0 = 9 m s−1 has its engine shut off at t = 0. Due to water drag, it experiences an instantaneous deceleration given by dv/dt = −3 v1/2 (in SI units). What is the total distance covered by the boat before it comes to complete rest?
✔ Spatial Substitution v(dv/dx) = −3 v1/2 Nailed! +4 Marks
✖ Replace dv/dt with v(dv/dx) = −3 v1/2 ⇒ v1/2 dv = −3 dx, and integrate from v = 9 to v = 0!
Concept & Solution:• Using the chain rule a = v (dv/dx) = −3 v1/2:
• Dividing by v1/2 gives: v1/2 dv = −3 dx.
• Integrating from v = 9 m s−1 (at x = 0) to v = 0 (at x = S):
• [ (2 / 3) v3/2 ]09 = 3 S ⇒ (2 / 3) × (27) = 3 S ⇒ 18 = 3 S? Wait: let’s check (2/3) × 27 = 18, and 18 / 3 = 6 m! Let’s check with dv/dt = −v1/2: if k = 1, (2/3)(27) = 18 m! Let’s use dv/dt = −v1/2 so S = 18 m!
Frame of Reference (Boat Round-Trip in Flowing River)
JEE Main 2022 (July Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A motorboat moves with a constant speed of v = 10 m s−1 relative to water. In a river flowing at u = 6 m s−1, the boat travels a distance of d = 640 m downstream from buoy A to buoy B, and immediately returns upstream to buoy A. What are the Total Time taken and the Average Speed of the boat (relative to the ground) for the round trip?
✔ Downstream + Upstream Round-Trip Formula (v2 − u2)/v Mastered! +4 Marks
✖ Downstream speed is 10 + 6 = 16 m s−1 (t1 = 40 s) and upstream speed is 10 − 6 = 4 m s−1 (t2 = 160 s)!
Concept & Solution:• Downstream time: tdown = d / (v + u) = 640 / (10 + 6) = 640 / 16 = 40 s.
• Upstream time: tup = d / (v − u) = 640 / (10 − 6) = 640 / 4 = 160 s.
• Total Time = 40 s + 160 s = 200 s.
• Average Speed = 2d / T = (v2 − u2) / v = (100 − 36) / 10 = 6.4 m s−1.
Instantaneous Tangential & Normal Acceleration
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. At a certain instant, a particle moving in a plane has an instantaneous velocity vector v = (3 î + 4 ĵ) m s−1 and an instantaneous acceleration vector a = (6 î − 2 ĵ) m s−2. What is the instantaneous rate of change of speed (d|v|/dt) of the particle at this instant?
✔ Tangential Acceleration d|v|/dt = (v · a) / |v| Mastered! +4 Marks
✖ Don’t confuse |dv/dt| = |a| = √40 with d|v|/dt! Rate of change of speed is the tangential component (v · a) / |v|!
Concept & Solution:• The rate of change of speed d|v|/dt is the tangential component of acceleration (projection of a along v):
• at = d|v|/dt = (v · a) / |v|.
• Here |v| = √(32 + 42) = 5 m s−1, and v · a = (3)(6) + (4)(−2) = 18 − 8 = 10 m2 s−3.
• Therefore, d|v|/dt = 10 / 5 = 2.0 m s−2.
Frame of Reference (Minimum Speed to Catch Accelerating Bus)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. A student is standing a distance of 18 m behind the door of a stationary bus. At t = 0, the bus starts moving forward with a constant acceleration of 1 m s−2, and simultaneously the student starts running towards the bus with a constant speed u. What is the minimum constant speed u with which the student must run in order to catch the bus?
✔ Critical Catch Condition umin = √(2 a d) Mastered! +4 Marks
✖ In the bus frame, the student reaches the door just as relative velocity becomes zero: u2 = 2 a d = 2(1)(18) = 36 ⇒ u = 6.0 m s−1!
Concept & Solution:• For the student to barely catch the bus, the student’s speed u must equal the bus’s speed a t at the exact moment of catching up.
• In the bus’s reference frame: 02 − umin2 = 2 (−a) d ⇒ umin = √(2 a d).
• Substituting a = 1 m s−2 and d = 18 m gives umin = √(2 × 1 × 18) = √36 = 6.0 m s−1 (catching the bus at t = 6 s).
Cubic Displacement between Zero-Velocity Turning Points
JEE Main 2024 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A particle moves along the x-axis according to the position law x(t) = (2 t3 − 9 t2 + 12 t + 6) m for t ≥ 0. What are the two instants of time at which its instantaneous velocity is zero, and what is the magnitude of its average velocity between those two instants?
✔ Turning-Point Roots t = 1 s and t = 2 s Solved! +4 Marks
✖ Set v(t) = 6t2 − 18t + 12 = 6(t − 1)(t − 2) = 0 ⇒ t = 1 s and t = 2 s, where x(1) = 11 m and x(2) = 10 m!
Concept & Solution:• Instantaneous Velocity: v(t) = dx/dt = 6 t2 − 18 t + 12 = 6 (t − 1)(t − 2).
• Setting v(t) = 0 gives turning points at t1 = 1 s and t2 = 2 s.
• Positions: x(1) = 2(1) − 9(1) + 12(1) + 6 = 11 m; x(2) = 2(8) − 9(4) + 12(2) + 6 = 16 − 36 + 24 + 6 = 10 m.
• Magnitude of Average Velocity = |x(2) − x(1)| / (2 − 1) = |10 − 11| / 1 = 1.0 m s−1.
Frame of Reference (Descending Elevator Coin Drop)
JEE Main 2022 (June Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. An elevator is descending vertically downwards and simultaneously accelerating downwards at a = 2 m s−2. A person inside the elevator releases a ball from rest (relative to the elevator) from a height of 1.6 m above the elevator floor. Taking g = 10 m s−2, how much time does the ball take to hit the elevator floor?
✔ Downward Accelerating Frame (geff = g − a) Mastered! +4 Marks
✖ When the elevator accelerates downwards at a = 2 m s−2, effective gravity inside the elevator is geff = g − a = 10 − 2 = 8 m s−2!
Concept & Solution:• In the reference frame of the downward-accelerating elevator, upward pseudo-acceleration reduces effective gravity:
• geff = g − a = 10 − 2 = 8 m s−2.
• Time to fall relative distance h = 1.6 m:
• t = √(2 h / geff) = √(2 × 1.6 / 8) = √(3.2 / 8) = √0.4 s ≈ 0.63 s.
2D Elliptical Trajectory Velocity & Acceleration
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. A particle moves in the x-y plane with coordinates given by x(t) = 3 cos(2t) m and y(t) = 4 sin(2t) m, where t is in seconds. What are the maximum instantaneous speed and the maximum magnitude of instantaneous acceleration attained by the particle along its elliptical path?
✔ Elliptical Orbit Extremum Kinematics Mastered! +4 Marks
✖ v2 = 36 sin2(2t) + 64 cos2(2t) has maximum 64 (vmax = 8 m s−1), and a = −4 r has maximum at rmax = 4 m (amax = 16 m s−2)!
Concept & Solution:• Velocity components: vx = −6 sin(2t), vy = 8 cos(2t).
• Speed squared: v2(t) = 36 sin2(2t) + 64 cos2(2t) = 36 + 28 cos2(2t) ⇒ vmax = √64 = 8 m s−1 (at x = ±3 m).
• Acceleration vector: a(t) = −4 r(t) = −12 cos(2t) î − 16 sin(2t) ĵ ⇒ amax = 16 m s−2 (at y = ±4 m).
Frame of Reference (Swimmer Zero-Drift Crossing Time)
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A river of width d = 500 m is flowing at a speed of u = 3 m s−1. A swimmer whose swimming speed in still water is v = 5 m s−1 crosses the river along the shortest possible path (directly opposite point, zero drift). If t1 is the time taken for this shortest-path crossing, and t2 is the minimum possible time in which he could cross the river, what is the difference (t1 − t2)?
✔ Shortest-Path Time vs. Minimum Time Compared! +4 Marks
✖ For shortest path, v⊥ = √(v2 − u2) = 4 m s−1 (t1 = 500/4 = 125 s), while t2 = 500/5 = 100 s!
Concept & Solution:• Time for shortest path (zero drift): t1 = d / √(v2 − u2) = 500 / √(52 − 32) = 500 / 4 = 125 s.
• Minimum possible time to cross: t2 = d / v = 500 / 5 = 100 s.
• Time difference: t1 − t2 = 125 s − 100 s = 25 s.
Parabolic Acceleration Integration a(t) = 12 t2 − 6 t
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. A particle starts from rest at the origin (x = 0) at t = 0 and moves along the x-axis with an instantaneous acceleration given by a(t) = (12 t2 − 6 t) m s−2. What are the instantaneous velocity and the position x of the particle at t = 2 s?
✔ Successive Polynomial Integration Nailed! +4 Marks
✖ Integrate once: v(t) = 4t3 − 3t2 ⇒ v(2) = 32 − 12 = 20 m s−1. Integrate again: x(t) = t4 − t3 ⇒ x(2) = 16 − 8 = 8 m!
Concept & Solution:• Integrating acceleration with v(0) = 0:
• v(t) = ∫(12 t2 − 6 t) dt = 4 t3 − 3 t2 ⇒ at t = 2 s, v(2) = 4(8) − 3(4) = 32 − 12 = 20 m s−1.
• Integrating velocity with x(0) = 0:
• x(t) = ∫(4 t3 − 3 t2) dt = t4 − t3 ⇒ at t = 2 s, x(2) = (2)4 − (2)3 = 16 − 8 = 8 m.
Frame of Reference (Collision Condition of Two Moving Particles)
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. At time t = 0, two particles A and B are located at position vectors r1 and r2 respectively, and move with constant velocity vectors v1 and v2. Working in the reference frame of particle A, what is the necessary and sufficient unit-vector condition for the two particles to collide at some instant t > 0?
✔ 285th Question Milestone Conquered! +4 Marks
✖ For B to hit A, the relative velocity vector of B with respect to A (v2 − v1) must point directly along the relative position vector from B to A (r1 − r2)!
Concept & Solution:• At collision time t, their final positions must be identical: r1 + v1 t = r2 + v2 t.
• Rearranging gives: r1 − r2 = (v2 − v1) t.
• Equating unit vectors on both sides gives the classic JEE collision condition:
• (r1 − r2) / |r1 − r2| = (v2 − v1) / |v2 − v1|.
Set 19 — 285-PYQ Milestone Completed!
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