Case 1 — Round-trip walk
Ria walks 200 m east from her home to the park, then returns 200 m west back to her home.
Q1. What is the total distance Ria traveled?
Total distance = 200 + 200 = 400 m.
Q2. What is her net displacement?
Displacement = final position − initial position = 0 m (she is back at home).
Case 2 — Bus trip with stops
A bus travels along a straight road: 10 km at 60 km/h, stops for 15 minutes, then travels 5 km at 30 km/h.
Q3. Find the average speed for the entire trip (include stop time).
Time1 = 10/60 = 1/6 h ≈ 0.1667 h. Stop time = 0.25 h. Time2 = 5/30 = 1/6 h ≈ 0.1667 h. Total distance = 15 km. Total time ≈ 0.1667+0.25+0.1667 = 0.5834 h. Average speed = 15 / 0.5834 ≈ 25.71 km/h.
Q4. Is average velocity same as average speed here?
Since motion is along a straight line without returning, displacement = distance = 15 km, so average velocity magnitude = average speed (≈25.71 km/h).
Case 3 — Constant acceleration car
A car starts from rest and accelerates uniformly to 72 km/h in 10 s.
Q5. Convert final speed to m/s and compute acceleration.
72 km/h = 20 m/s. a = (v−u)/t = (20−0)/10 = 2 m/s².
Q6. How far did the car travel in this time?
s = ut + ½at² = 0 + 0.5×2×100 = 100 m.
Case 4 — Braking car
A car traveling at 20 m/s applies brakes and stops in 5 s with uniform deceleration.
Q7. Find deceleration and stopping distance.
a = (0−20)/5 = −4 m/s² (deceleration 4 m/s²). s = ut + ½at² = 20×5 + 0.5×(−4)×25 = 100 − 50 = 50 m.
Case 5 — Runner with changing pace
A runner covers first 100 m at constant 5 m/s and next 100 m at 10 m/s.
Q8. Compute total time and average speed for 200 m.
Time1 = 100/5 = 20 s; Time2 = 100/10 = 10 s; Total time = 30 s. Average speed = total distance/total time = 200/30 ≈ 6.67 m/s.
Case 6 — Vertical toss
A ball is thrown upward with 14 m/s. Take g = 10 m/s².
Q9. Time to reach highest point and maximum height?
t = u/g = 14/10 = 1.4 s. H = u²/(2g) = 196/(20) = 9.8 m.
Q10. Total time of flight (neglect air resistance)?
Total time = 2 × 1.4 = 2.8 s.
Case 7 — Free fall from tower
A stone is dropped from rest from a tower and hits ground in 3 s (g = 9.8 m/s²).
Q11. Height of tower and impact speed?
s = ½ g t² = 0.5×9.8×9 = 44.1 m. v = gt = 9.8×3 = 29.4 m/s (downwards).
Case 8 — Crossing trains (relative motion)
Two trains, 150 m and 200 m long, move towards each other at 54 km/h and 36 km/h respectively on parallel tracks.
Q12. How long to pass each other completely?
Relative speed = 54+36 = 90 km/h = 25 m/s (90×5/18). Total length = 350 m. Time = 350 / 25 = 14 s.
Case 9 — Boat in river
A boat's speed in still water is 8 km/h. River current is 3 km/h.
Q13. Speed upstream and downstream?
Upstream = 8 − 3 = 5 km/h. Downstream = 8 + 3 = 11 km/h.
Q14. If it travels 22 km downstream, time taken?
Time = distance / speed = 22 / 11 = 2 h.
Case 10 — Motion graphs
A v–t graph shows velocity = 0 m/s for t = 0–2 s, then constant 6 m/s for t = 2–5 s, then decreases linearly to 0 at t = 8 s.
Q15. Displacement from 0 to 8 s?
Area: 0–2 s → 0. 2–5 s rectangle: 6×3 = 18 m. 5–8 s triangle: base 3, height 6 → area = 0.5×3×6 = 9 m. Total = 27 m.
Q16. Interpret acceleration in 5–8 s interval.
Velocity decreases linearly from 6 to 0 in 3 s → acceleration = (0−6)/3 = −2 m/s² (uniform deceleration).
Case 11 — Two-stage motion
A cyclist accelerates uniformly from 2 m/s to 8 m/s in 6 s, then travels at constant 8 m/s for 10 s.
Q17. Acceleration during first stage and total displacement in 16 s?
a = (8−2)/6 = 1 m/s². Displacement stage1: s1 = ut + ½at² = 2×6 + 0.5×1×36 = 12 + 18 = 30 m. Stage2: s2 = v×t = 8×10 = 80 m. Total = 110 m.
Case 12 — Stopping sight distance
A vehicle at 25 m/s has a driver's reaction time of 0.8 s and brakes decelerate at 5 m/s².
Q18. Compute reaction distance, braking distance and stopping distance.
Reaction distance = speed × reaction time = 25 × 0.8 = 20 m. Braking distance using v² = u² + 2as (v=0): s = u²/(2a) = 25²/(2×5) = 625/10 = 62.5 m. Total stopping distance = 20 + 62.5 = 82.5 m.
Case 13 — Motion in two parts (use v²)
A car traveling at 30 m/s decelerates uniformly to 10 m/s while covering 80 m.
Q19. Find deceleration using v² = u² + 2as.
10² = 30² + 2 a (80) → 100 = 900 + 160 a → 160 a = −800 → a = −5 m/s².
Case 14 — Motion with zero average velocity
A toy car moves in a straight line from A to B and back to A in 20 s covering 40 m total.
Q20. What are average speed and average velocity?
Average speed = total distance/total time = 40/20 = 2 m/s. Average velocity = net displacement/time = 0/20 = 0 m/s (since returns to start).
Case 15 — Projectile horizontal launch
A small stone is projected horizontally from a cliff with speed 12 m/s and lands 4 s later.
Q21. Horizontal range and height of cliff?
Horizontal range = 12 × 4 = 48 m. Height = ½ g t² ≈ 0.5×9.8×16 = 78.4 m.
Case 16 — Changing reference frame
A man walks at 4 m/s on a train that moves at 10 m/s relative to ground in same direction.
Q22. What is man's speed relative to ground? If train direction reverses and man still walks same direction relative to train, what is his ground speed?
If man's direction same as train: v_ground = 10 + 4 = 14 m/s. If train reverses direction (train velocity −10 m/s relative to previous frame) and man still walks at 4 m/s relative to train in the original walking direction, relative ground speed = (−10) + 4 = −6 m/s (i.e., 6 m/s opposite original positive direction).
Case 17 — Non-uniform acceleration (linear v(t))
A particle moves so that v(t) = 2t + 1 (m/s). Time in seconds.
Q23. Find acceleration (function of t) and displacement from t=0 to t=3 s.
a(t) = dv/dt = 2 m/s² (constant). Displacement = ∫0→3 v dt = ∫0→3 (2t +1) dt = [t² + t]0→3 = 9 + 3 = 12 m.
Case 18 — Two runners chasing
Runner A at 6 m/s chases Runner B at 4 m/s; when does A catch B if initial gap is 40 m?
Q24. Time to catch and distance covered by A in that time?
Relative speed = 6 − 4 = 2 m/s. Time = distance/gap ÷ relative speed = 40 / 2 = 20 s. Distance by A = 6 × 20 = 120 m.
Case 19 — Acceleration sign convention
A ball moves along +x direction with velocity +10 m/s and slows to +2 m/s in 4 s.
Q25. What is acceleration and its sign? If ball reverses direction to −3 m/s afterwards in 1 s, what is acceleration in that interval?
First interval: a = (2−10)/4 = −8/4 = −2 m/s² (negative → deceleration). Second interval: a = (−3 − 2)/1 = −5 m/s² (large negative acceleration causing reversal).
Case 20 — Exam strategy case
You are solving a kinematics question in exam: known u, s and a; asked for t.
Q26. Which equation will you use and how will you pick the physically valid root?
Use s = ut + ½ a t² → rearrange to ½ a t² + u t − s = 0 (quadratic in t). Solve using quadratic formula. Choose t ≥ 0 and appropriate based on context (ignore negative root unless time before t=0 has physical meaning). Convert units consistently and check result by plugging back into equations.
Q27. Quick checklist before finalizing numeric answer?
Check units, sign convention, whether magnitude is reasonable, and substitute into an independent relation (like v = u + at if possible) to verify consistency.