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Distance, displacement, speed, velocity.
Equations of motion and free fall.
Graphical analysis of motion.
Relative motion and applications.
A comprehensive mock covering all subtopics — displacement and velocity, equations of motion, free fall, graphical analysis, and relative motion. Ideal after reading the notes and working through solved examples.
1. A car travels 4 km east and then 3 km north. The displacement is:
Explanation: Displacement = √(4² + 3²) = √25 = 5 km . Distance = 7 km (path length).
2. A particle covers equal distances in equal intervals of time. This type of motion is called:
Explanation: Uniform motion is defined as motion in which a body covers equal distances in equal intervals of time.
3. If the velocity of a body is constant, its acceleration is:
Explanation: Acceleration = rate of change of velocity. If velocity is constant, dv/dt = 0, so acceleration is zero .
4. The first equation of motion is:
Explanation: The first equation of uniform acceleration is v = u + at , relating final velocity to initial velocity, acceleration, and time.
5. A body is dropped from rest. The velocity after 3 seconds (g = 10 m/s²) is:
Explanation: u = 0, a = g = 10 m/s², t = 3 s. v = u + at = 0 + 10×3 = 30 m/s .
6. A person walks 5 m east, then 5 m west. The total distance and displacement are:
Explanation: Distance = 5 + 5 = 10 m (total path). Displacement = 5 − 5 = 0 m (net position change).
7. Average speed is defined as:
Explanation: Average speed = total distance / total time . It is a scalar. Average velocity = total displacement / total time.
8. In a position-time graph, a horizontal straight line indicates:
Explanation: A horizontal line on a position-time graph means position is constant. The body is at rest . Slope = velocity = 0.
9. A car accelerates from 0 to 20 m/s in 5 seconds. Its acceleration is:
Explanation: a = (v−u)/t = (20−0)/5 = 4 m/s² .
10. A stone is thrown vertically upward. At the highest point, its velocity is:
Explanation: At the highest point , the stone momentarily stops before returning. Its velocity is zero , though acceleration due to gravity still acts downward.
11. Displacement can be:
Explanation: Displacement is a vector and can be zero, positive, or negative , depending on direction. Distance is always non-negative.
12. Negative acceleration is also called:
Explanation: When acceleration is negative (opposite to velocity), the body slows down. This is called deceleration or retardation .
13. The distance covered in the nth second of uniform acceleration is given by:
Explanation: Distance in nth second: sₙ = u + a(2n−1)/2 . Derived from sₙ − sₙ₋₁ using the second equation of motion.
14. In a velocity-time graph, a straight line with positive slope represents:
Explanation: Slope of v-t graph = acceleration. A straight line with positive slope means constant positive acceleration — uniform acceleration .
15. Two bodies of different masses are dropped simultaneously from the same height in vacuum. They reach the ground:
Explanation: In vacuum, all bodies fall with the same acceleration g. They reach the ground simultaneously (Galileo's law).
16. A train starts from rest and accelerates uniformly at 2 m/s². The distance covered in 10 seconds is:
Explanation: s = ut + ½at² = 0 + ½(2)(100) = 100 m .
17. A body moving at 30 m/s is decelerated at 5 m/s². The distance covered before stopping is:
Explanation: v² = u² − 2as. At rest: 0 = 900 − 10s → s = 90 m .
18. A ball is thrown vertically upward with velocity 20 m/s (g = 10 m/s²). The maximum height reached is:
Explanation: v² = u² − 2gh. At max height: 0 = 400 − 20h → h = 20 m .
19. A train moves at 60 km/h and a car moves at 40 km/h in the same direction. The velocity of the car relative to the train is:
Explanation: Relative velocity of car w.r.t. train = 40 − 60 = −20 km/h . The negative sign means the car moves backward relative to the train.
20. The area under a speed-time graph represents:
Explanation: Area under speed-time graph = ∫v dt = distance covered .
21. A particle starts from rest and moves with acceleration a = 4 m/s². The velocity at the end of 5th second is:
Explanation: v = u + at = 0 + 4×5 = 20 m/s .
22. Two trains (200 m and 300 m long) run on parallel tracks in opposite directions at 40 m/s and 60 m/s. Time to completely cross each other is:
Explanation: Relative speed = 40 + 60 = 100 m/s. Total length = 500 m. Time = 500/100 = 5 s .
23. A stone dropped from the top of a building takes 4 seconds to reach the ground (g = 10 m/s²). The height of the building is:
Explanation: h = ut + ½gt² = 0 + ½(10)(16) = 80 m .
24. If the position-time graph of a particle is a curve (not straight line), the particle has:
Explanation: A curve on the x-t graph means the slope (velocity) is changing, implying non-uniform velocity and acceleration.
25. A train running at 72 km/h is stopped by brakes in 8 seconds. The retardation is:
Explanation: u = 72 km/h = 20 m/s, v = 0, t = 8 s. a = (0−20)/8 = −2.5 m/s². Retardation = 2.5 m/s² .
26. A swimmer can swim at 3 m/s in still water. A river flows at 4 m/s. The swimmer swims perpendicular to the flow. The resultant speed is:
Explanation: Resultant = √(3² + 4²) = √25 = 5 m/s .
27. A stone is thrown downward with velocity 10 m/s from a height of 80 m (g = 10 m/s²). Time to reach ground:
Explanation: Using $s=ut+\frac12gt^2$ gives $80=10t+5t^2$, or $t^2+2t-16=0$. Its positive root is $t=-1+√17= (√17-1)\text{ s} $, about 3.12 s.
28. For uniform motion, the position-time (x-t) graph is:
Explanation: For uniform motion (constant velocity), x = x₀ + vt is linear. The x-t graph is a straight line with nonzero slope .
29. A bullet fired from a gun has muzzle velocity 500 m/s and decelerates at 100 m/s² in a sand bag. Distance penetrated before stopping:
Explanation: v² = u² − 2as → 0 = 250000 − 200s → s = 1250 m .
30. Two cars A and B start from the same point. A goes north at 60 km/h, B goes east at 80 km/h. After 1 hour the distance between them is:
Explanation: After 1 h: A is 60 km north, B is 80 km east. Distance = √(60²+80²) = 100 km .
31. A particle moves along x-axis with acceleration a = 2t m/s². If v = 0 at t = 0, the velocity at t = 3 s is:
Explanation: v = ∫a dt = ∫2t dt = t² + C. At t=0, v=0 → C=0. At t=3: v = 9 = 9 m/s .
32. A particle's position is x = 3t³ − 2t² + t. The acceleration at t = 2 s is:
Explanation: v = dx/dt = 9t² − 4t + 1. a = dv/dt = 18t − 4. At t=2: a = 36−4 = 32 m/s² .
33. The velocity-time graph of a particle is a straight line making 45° with the time-axis. If the particle starts from rest, the displacement in 4 s is:
Explanation: Acceleration = tan45° = 1 m/s². s = ½at² = ½(1)(16) = 8 m .
34. A train of length 200 m passes a platform of length 300 m at 10 m/s. The time taken is:
Explanation: Distance to cross = train length + platform length = 200 + 300 = 500 m at 10 m/s. Time = 500/10 = 50 s .
35. A particle at rest starts moving with acceleration a = (4 − t) m/s². At what time does the velocity become maximum?
Explanation: Velocity is maximum when a = 0: 4 − t = 0 → t = 4 s . After this, a becomes negative and velocity decreases.
36. A particle's x-t graph is an upward-opening parabola. This means the particle has:
Explanation: x = ut + ½at² is a parabola. An upward-opening parabola indicates constant positive acceleration .
37. A ball is dropped from a height H. After falling through H/2, its speed is:
Explanation: v² = 2g(H/2) = gH → v = √(gH) .
38. In the nth second, a body starting from rest covers a distance of (2n − 1) units. The acceleration is:
Explanation: Distance in nth second: sₙ = u + a(2n−1)/2 = 0 + a(2n−1)/2. Given sₙ = 2n−1: a/2 = 1 → a = 2 unit/s² .
39. The velocity of a particle is v = √(4 + 4x), where x is in metres. The acceleration is:
Explanation: a = v(dv/dx). v² = 4 + 4x → 2v(dv/dx) = 4 → v(dv/dx) = 2. So a = 2 m/s² (constant).
40. The displacement of a particle is s = 6 + 12t − 2t². The particle comes to rest at t =
Explanation: v = ds/dt = 12 − 4t. Setting v = 0: 12 − 4t = 0 → t = 3 s .
41. The ratio of distances covered in 1st, 2nd, and 3rd seconds by a body starting from rest under uniform acceleration is:
Explanation: Distance in nth second ∝ (2n−1). For n=1,2,3: (2×1−1):(2×2−1):(2×3−1) = 1:3:5 .
42. A particle starts with velocity 10 m/s and has constant deceleration 2 m/s². The total distance before it stops and returns to start is:
Explanation: Distance to stop: s₁ = v²/2a = 100/4 = 25 m. On return it covers another 25 m. Total = 50 m .
43. A rocket is fired vertically with net upward acceleration a for time T. After T, it goes into free fall. The maximum height above ground is:
Explanation: Height at burnout: h₁ = ½aT². Velocity at burnout: v = aT. Extra height: h₂ = v²/2g = a²T²/2g. Total max height = ½aT² + a²T²/2g .
44. A ball is thrown vertically upward with velocity 30 m/s from the top of a 45 m tower (g = 10 m/s²). The speed when it hits the ground is:
Explanation: By $v^2=u^2+2as$, taking upward as positive gives $v^2=30^2+2(-10)(-45)=1800$. Therefore the impact speed is $√1800= 30√2\text{ m/s} $.
45. The displacement of a particle varies with time as x = 4t − t² (in SI units). When does the particle return to the origin?
Explanation: x = 0: 4t − t² = 0 → t(4−t) = 0 → t = 0 or t = 4 s . At t = 0 (start) and t = 4 s (returns to origin).
46. A cyclist covers 200 m in 20 s. His average speed is:
Explanation: Average speed = 200/20 = 10 m/s .
47. A body accelerates from rest to 10 m/s in 5 s, then decelerates to rest in 10 s. The total distance is:
Explanation: Phase 1: s₁ = ½(0+10)(5) = 25 m. Phase 2: s₂ = ½(10+0)(10) = 50 m. Total = 75 m .
48. A person walks at 4 km/h for 2 hours and then at 6 km/h for 3 hours. His average speed is:
Explanation: Total distance = 8 + 18 = 26 km. Total time = 5 h. Average speed = 26/5 = 5.2 km/h .
49. A body thrown vertically up passes a point P at height h with velocity v. The maximum height reached above P is:
Explanation: At P, velocity = v (upward). Extra height above P: using v² = 2gH → H = v²/2g .
50. Two cars start from same point. Car A at 40 km/h, car B at 60 km/h in same direction. After 2 hours separation is:
Explanation: A covers 80 km, B covers 120 km. Separation = 120 − 80 = 40 km .
51. A stone is dropped into a well. The sound of splash is heard after 4 seconds. If speed of sound = 340 m/s and g = 10 m/s², the depth of the well is approximately:
Explanation: d ≈ 75 m: t₁ = √(2×75/10) = √15 ≈ 3.87 s, t₂ = 75/340 ≈ 0.22 s. Total ≈ 4.09 s ≈ 4 s. Answer: 75 m .
52. The braking distance of a vehicle moving at speed v with deceleration a is:
Explanation: v² = u² − 2as. At rest: 0 = v² − 2a·s → s = v²/2a .
53. The standard acceleration due to gravity g is approximately:
Explanation: The standard value is g ≈ 9.8 m/s² (approximated as 10 m/s² in problems).
54. A body falling freely from height H has velocity v at height H/4 from ground. Then v is:
Explanation: Body falls from H to H/4, covering 3H/4. v² = 2g(3H/4) = 3gH/2 → v = √(3gH/2) .
55. If a body covers S₁ in time t and S₂ in time 2t with uniform acceleration starting from rest, then S₁:S₂ is:
Explanation: S = ½at². S₁ = ½at², S₂ = ½a(2t)² = 4×½at². S₁:S₂ = 1:4 .
56. The displacement of a particle in time t is given by s = at + bt². The initial velocity and acceleration are:
Explanation: Comparing with s = ut + ½(Acc)t²: u = a (initial velocity), ½ × Acc = b → Acc = 2b . So initial velocity = a, acceleration = 2b.
57. A bullet of mass 10 g traveling at 200 m/s is stopped in 0.05 s. The retarding force is:
Explanation: F = m(Δv/Δt) = 0.01 × (200/0.05) = 0.01 × 4000 = 40 N .
58. Which represents uniform motion on a position-time graph?
Explanation: Uniform motion means constant velocity. x = x₀ + vt is linear — a straight line with constant nonzero slope .
59. A car decelerates from 20 m/s to rest in 10 s. The deceleration is:
Explanation: a = (v−u)/t = (0−20)/10 = −2 m/s². Deceleration = 2 m/s² .
60. If body A is dropped from height h and body B is thrown horizontally from the same height, both will reach ground:
Explanation: Vertical motion of both is identical: free fall with initial vertical velocity = 0. Both reach ground simultaneously after time t = √(2h/g).