TOPIC: Frame of Reference, Distance, Displacement, Average & Instantaneous Velocity/Acceleration
IIT-JEE MAINS PYQ Arena — Set 18
Frame of Reference, Distance, Displacement & Avg/Instantaneous Kinematics
Set 18 • Kinematics IV
Instantaneous Velocity in Y-Z Plane
JEE Main 2025 (Jan Shift 1)
⏱ JEE Standard Pace: 120s
+4 / -1 Scale
Q1. The position vector of an ant (S in metres) moving in the Y-Z plane is given by S(t) = 2 t2 ĵ + 5 k̂ (where t is in seconds). What are the magnitude and direction of the instantaneous velocity of the ant at t = 1 s?
✔ Spot On! Derivative of Constant 5 k̂ is Zero! +4 Marks
✖ Differentiate S(t) = 2t2 ĵ + 5 k̂ with respect to t: v(t) = 4t ĵ, which gives 4 m s−1 in the y-direction at t = 1 s!
Concept & Solution:• Instantaneous Velocity vector: v(t) = dS/dt = d/dt (2 t2 ĵ + 5 k̂) = 4 t ĵ m s−1.
• At t = 1 s: v(1) = 4 ĵ m s−1, which has a magnitude of 4 m s−1 directed along the y-axis.
Question 1 of 15
Frame of Reference (Balloon & Dropped Stone Separation)
JEE Main 2026 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q2. A gas balloon is rising vertically upwards with a constant velocity of 10 m s−1. When the balloon reaches a height of 75 m above the ground, a stone is gently dropped from it while the balloon continues moving upwards at the same constant velocity of 10 m s−1. Taking g = 10 m s−2, what is the height of the balloon above the ground at the instant when the stone hits the ground?
✔ Latest 2026 PYQ Nailed! +4 Marks
✖ The stone takes t = 5 s to reach the ground (−75 = 10t − 5t2), and in 5 s the balloon rises an additional 50 m!
Concept & Solution:• When released, the stone has initial upward velocity u = +10 m s−1.
• Displacement of stone to reach the ground: −75 = 10 t − (1/2)(10) t2 ⇒ t2 − 2t − 15 = 0 ⇒ (t − 5)(t + 3) = 0 ⇒ t = 5 s.
• In these 5 seconds, the balloon continues rising at 10 m s−1: hballoon = 75 m + (10 m s−1 × 5 s) = 125 m.
Rectilinear Acceleration Law a(x) = α − β x
JEE Main 2025 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q3. A streetcar moves rectilinearly from rest at Station A to come to rest at the next Station B with an instantaneous acceleration varying with distance x from Station A according to the law a(x) = α − β x, where α and β are positive constants. What are the distance between the two stations and the maximum velocity attained by the streetcar?
✔ Classic Variable-Acceleration Integration Mastered! +4 Marks
✖ Use v dv = (α − βx) dx ⇒ v2/2 = αx − βx2/2. Setting v = 0 gives x = 2α/β, and maximum velocity occurs where a = 0 (x = α/β)!
Concept & Solution:• Using v dv = (α − β x) dx and integrating from x = 0 (v = 0):
• v2 / 2 = α x − (1/2) β x2 ⇒ v2 = 2 α x − β x2.
• At Station B, the car comes to rest again (v = 0): 2 α x − β x2 = 0 ⇒ x = 2α / β.
• Maximum velocity occurs where acceleration a(x) = 0 ⇒ x = α / β, giving vmax2 = 2α(α/β) − β(α/β)2 = α2 / β ⇒ vmax = α / √β.
Cubic Position Law & Velocity at Zero Acceleration
JEE Main 2024 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q4. The position of a body moving along a straight line is given by x(t) = (−3 t3 + 18 t2 + 16 t) m, where t is in seconds. What is the instantaneous velocity of the body at the instant when its instantaneous acceleration becomes zero?
✔ Spot On! Zero Acceleration at t = 2 s gives v = 52 m s−1! +4 Marks
✖ Differentiate twice: v(t) = −9t2 + 36t + 16 and a(t) = −18t + 36 = 0 at t = 2 s!
Concept & Solution:• Instantaneous Velocity: v(t) = dx/dt = −9 t2 + 36 t + 16.
• Instantaneous Acceleration: a(t) = dv/dt = −18 t + 36.
• Setting a(t) = 0 ⇒ −18 t + 36 = 0 ⇒ t = 2 s.
• Velocity at t = 2 s: v(2) = −9(2)2 + 36(2) + 16 = −36 + 72 + 16 = 52 m s−1.
Frame of Reference (Cylinder Rolling off Moving Truck)
JEE Main 2025 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q5. A solid cylinder rolls off the back of a large truck travelling horizontally at 10 m s−1 to the right. At the instant it leaves the truck, the cylinder is moving horizontally at 8 m s−1 to the left relative to an observer in the truck. If the cylinder lands on the roadway 1.25 m below its starting level (take g = 10 m s−2), how far behind the rear of the truck does it land?
✔ Truck Reference Frame Simplification Mastered! +4 Marks
✖ Work directly in the truck’s inertial frame! Time of fall t = √(2 × 1.25 / 10) = 0.5 s, and relative horizontal speed is 8 m s−1 backwards!
Concept & Solution:• Time taken to fall vertically by h = 1.25 m: t = √(2h / g) = √(2 × 1.25 / 10) = √0.25 = 0.5 s.
• Since the truck moves at constant velocity, its frame is inertial with zero horizontal acceleration.
• Distance behind the truck = |vrel,x| × t = 8 m s−1 × 0.5 s = 4.0 m.
Space-Average vs. Time-Average Velocity
JEE Main 2023 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q6. A particle starts from rest at x = 0 and moves with a constant acceleration a along the positive x-axis until it reaches an instantaneous velocity v0 at position x = L. What is the ratio of its Time-Average Velocity (⟨v⟩t = ∫v dt / ∫dt) to its Space-Average Velocity (⟨v⟩x = ∫v dx / ∫dx) over this interval?
✔ Time-Average (v0/2) vs. Space-Average (2v0/3) Mastered! +4 Marks
✖ Time-average velocity is v0/2, whereas Space-average velocity for v ∝ x1/2 is (2/3)v0, giving ratio (1/2) / (2/3) = 3 : 4!
Concept & Solution:• Time-Average Velocity for uniform acceleration from rest: ⟨v⟩t = (0 + v0) / 2 = v0 / 2.
• Since v(x) = √(2a x) = v0 (x / L)1/2, the Space-Average Velocity is:
• ⟨v⟩x = (1 / L) ∫0L v0 (x / L)1/2 dx = (2 / 3) v0.
• Ratio ⟨v⟩t / ⟨v⟩x = (v0 / 2) / (2 v0 / 3) = 3 : 4.
Frame of Reference (Non-Uniform River Velocity Profile)
JEE Main 2025 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q7. A swimmer can swim with speed u with respect to water. He crosses a river of width d in the shortest possible time (by swimming perpendicular to the banks). The speed of river water is zero at both banks and increases linearly to a maximum value v0 at the mid-stream of the river, then decreases linearly back to zero at the opposite bank. What is the total downstream drift of the swimmer?
✔ Linear Stream Profile Integration Nailed! +4 Marks
✖ Since river speed varies linearly from 0 to v0 and back to 0, the average river speed is v0/2, and crossing time is d/u!
Concept & Solution:• Since the swimmer swims perpendicular to the flow at constant speed u, total crossing time is T = d / u.
• Because the river speed profile is a triangle with base T and peak height v0, the average downstream speed of the water is vr,avg = v0 / 2.
• Total Downstream Drift = vr,avg × T = (v0 / 2) × (d / u) = (v0 d) / (2 u).
2D Instantaneous Velocity Angle with Horizontal
JEE Main 2026 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q8. If the x and y coordinates of a particle moving in a vertical plane as a function of time t (in seconds) are given as x(t) = 24 t and y(t) = 43.6 t − 4.9 t2 (in metres), then what is the angle made by the instantaneous velocity vector of the particle with the horizontal at t = 2 s?
✔ Latest 2026 Parametric Velocity Question Solved! +4 Marks
✖ Differentiate x(t) and y(t): at t = 2 s, vx = 24 m s−1 and vy = 43.6 − 9.8(2) = 24 m s−1, so tan θ = 1 ⇒ θ = 45°!
Concept & Solution:• Horizontal velocity component: vx = dx/dt = 24 m s−1.
• Vertical velocity component: vy = dy/dt = 43.6 − 9.8 t.
• At t = 2 s: vy(2) = 43.6 − 9.8(2) = 43.6 − 19.6 = 24 m s−1.
• Angle with horizontal: tan θ = vy / vx = 24 / 24 = 1 ⇒ θ = 45°.
Frame of Reference (Two Cars Starting from Rest)
JEE Main 2023 (Jan Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q9. A car is standing 200 m behind a bus, which is also at rest on a straight road. Both start moving forward at the exact same instant (t = 0), with the bus accelerating at aB = 2 m s−2 and the car accelerating at aC = 4 m s−2. After how much time will the car catch up with the bus, and what is the speed of the car relative to the bus at that instant?
✔ Relative Acceleration srel = (1/2) arel t2 Mastered! +4 Marks
✖ Relative acceleration aCB = 4 − 2 = 2 m s−2, so 200 = (1/2)(2) t2 ⇒ t = √200 = 10√2 s!
Concept & Solution:• Initial relative velocity: uCB = 0.
• Relative acceleration of car with respect to bus: aCB = aC − aB = 4 − 2 = 2 m s−2.
• Using sCB = (1/2) aCB t2 ⇒ 200 = (1/2)(2) t2 ⇒ t2 = 200 ⇒ t = 10√2 s.
• Relative speed at catch-up: vCB = aCB t = 2 × 10√2 = 20√2 m s−1.
Instantaneous Acceleration from v(x) = 4 √x
JEE Main 2024 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q10. A particle initially at rest starts moving from reference point x = 0 along the positive x-axis with an instantaneous velocity v that varies with position x as v(x) = 4 √x m s−1. What are the instantaneous acceleration of the particle and its average velocity during the first 3 seconds of motion?
✔ Squaring v2 = 16x → a = 8 m s−2 Nailed! +4 Marks
✖ Square both sides: v2 = 16x, and compare with v2 = 2ax to get a = 8 m s−2!
Concept & Solution:• Squaring the velocity relation: v2 = 16 x.
• Differentiating with respect to x: 2v (dv/dx) = 16 ⇒ a = v (dv/dx) = 8 m s−2 (constant acceleration).
• For constant acceleration from rest over t = 3 s: final velocity v(3) = a t = 8 × 3 = 24 m s−1.
• Average Velocity = (0 + 24) / 2 = 12 m s−1.
Frame of Reference (Relative Speed of Stone Thrown Between Cars)
JEE Main 2026 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q11. Two cars A and B are moving in the same direction along a straight highway. Car A (ahead) moves at 100 km h−1 and Car B (behind) moves at 80 km h−1. A person sitting in Car B throws a stone forward with a speed v (relative to Car B) so that it hits Car A with a relative impact speed of 5 m s−1. Neglecting vertical motion and air resistance, what is the value of v in km h−1?
✔ Latest 2026 Relative Velocity Conversion Nailed! +4 Marks
✖ Convert 5 m s−1 to 18 km h−1 first! Then (80 + v) − 100 = 18 ⇒ v = 38 km h−1!
Concept & Solution:• Impact speed relative to Car A: 5 m s−1 = 5 × (18 / 5) = 18 km h−1.
• Speed of the stone with respect to the ground: vstone,G = 80 + v (in km h−1).
• Speed of stone relative to Car A: vstone,A = vstone,G − vA = (80 + v) − 100 = 18.
• Therefore, v − 20 = 18 ⇒ v = 38 km h−1.
Distance vs. Displacement in Circular Arc (Angle θ)
JEE Main 2023 (Apr Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q12. A particle moves along a circular path of radius R with constant speed v. When the radius vector of the particle sweeps through an angle of θ = 60° (π / 3 rad), what are the magnitude of its displacement (|Δr|) and the magnitude of the change in its velocity vector (|Δv|)?
✔ Equilateral Vector Triangle 2 sin(θ/2) = 1 Mastered! +4 Marks
✖ Use the chord formula |Δr| = 2 R sin(θ/2) and |Δv| = 2 v sin(θ/2). For θ = 60°, 2 sin(30°) = 1!
Concept & Solution:• Magnitude of Displacement (chord length): |Δr| = 2 R sin(θ / 2) = 2 R sin(30°) = R (while path distance = πR/3).
• Magnitude of change in velocity: |Δv| = 2 v sin(θ / 2) = 2 v sin(30°) = v.
2D Constant Acceleration Speed Calculation
JEE Main 2024 (Jan Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q13. A particle moves in the x-y plane with a constant acceleration of a = (1 î − 1 ĵ) m s−2. If its initial velocity at t = 0 is u = (2 î + 3 ĵ) m s−1, what are the instantaneous speed of the particle at t = 2 s and the magnitude of its displacement during these 2 seconds?
✔ 2D Vector Equations of Motion Applied Perfectly! +4 Marks
✖ At t = 2 s, v = (2+2)î + (3−2)ĵ = 4î + 1ĵ m s−1 (speed = √17), and Δr = 6î + 4ĵ m (|Δr| = √52 = 2√13 m)!
Concept & Solution:• Velocity at t = 2 s: v = u + a t = (2î + 3ĵ) + 2(1î − 1ĵ) = 4 î + 1 ĵ m s−1.
• Instantaneous Speed: |v| = √(42 + 12) = √17 m s−1.
• Displacement in 2 s: Δr = [(u + v) / 2] × t = [(6î + 4ĵ) / 2] × 2 = 6 î + 4 ĵ m.
• Magnitude of Displacement: |Δr| = √(62 + 42) = √52 = 2√13 m.
Frame of Reference (Same vs. Opposite Direction Train Crossing Ratio)
JEE Main 2022 (July Shift 2)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q14. Two trains of lengths 60 m and 120 m move on parallel tracks with constant speeds of 80 km h−1 and 30 km h−1 respectively. If t1 is the time taken for them to completely cross each other when moving in the same direction, and t2 is the time taken when moving in opposite directions, what is the ratio (t1 / t2)?
✔ Inverse Relative Speed Ratio (v₁ + v₂)/(v₁ − v₂) Nailed! +4 Marks
✖ Since total length (180 m) cancels out, t₁ / t₂ = (v₁ + v₂) / (v₁ − v₂) = (80 + 30) / (80 − 30) = 110 / 50 = 11 / 5!
Concept & Solution:• Total relative distance to cover in both cases: Ltotal = 60 m + 120 m = 180 m.
• Same direction: t1 = Ltotal / (v1 − v2) = 180 / (80 − 30) = 180 / 50.
• Opposite direction: t2 = Ltotal / (v1 + v2) = 180 / (80 + 30) = 180 / 110.
• Ratio t1 / t2 = (v1 + v2) / (v1 − v2) = 110 / 50 = 11 : 5.
Instantaneous Speed Equal to Numerical Value of g
JEE Main 2026 (Apr Shift 1)
⏱ JEE Standard Pace: 120s+4 / -1 Scale
Q15. From a height of 18 m above the ground, a ball is dropped from rest. Neglecting air resistance and taking g = 10 m s−2, at what height above the ground does the numerical value of its instantaneous velocity become equal to the numerical value of its instantaneous acceleration (in SI units)?
✔ 270th Question Milestone Conquered! +4 Marks
✖ To reach speed v = 10 m s−1, the ball falls by hfall = v2/(2g) = 100/20 = 5 m from the top, so height above ground is 18 − 5 = 13 m!
Concept & Solution:• Numerical value of acceleration due to gravity: |a| = 10 m s−2, so we need the point where instantaneous speed v = 10 m s−1.
• Distance fallen from rest to reach v = 10 m s−1: hfall = v2 / (2g) = 102 / (2 × 10) = 5 m.
• Height above the ground = Total Height − hfall = 18 m − 5 m = 13 m.
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