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Distance, displacement, speed, and velocity fundamentals.
Equations of motion, free fall, and nth second formula.
Variable acceleration, calculus-based kinematics.
Graph analysis — s-t, v-t, and a-t graphs.
A comprehensive mock covering all subtopics — distance/velocity basics, suvat equations, free fall, variable acceleration, and graphs. Ideal after reading the notes and working through solved examples.
1. A particle moves 40 m north and then 30 m east. Its displacement magnitude is:
Explanation: Displacement = $\sqrt{40^2 + 30^2} = \sqrt{1600+900} = \sqrt{2500} = 50$ m. Distance = 70 m (sum of path lengths), but displacement is the straight-line distance.
2. A car travels 3 km east, then turns and goes 4 km north. The ratio of distance to displacement magnitude is:
Explanation: Distance = 3 + 4 = 7 km. Displacement = $\sqrt{9+16} = 5$ km. Ratio = 7/5.
3. A particle completes one full revolution around a circle of radius 7 m. Its displacement is:
Explanation: After one complete revolution the particle returns to its starting point. Displacement = final position − initial position = 0. Distance = circumference = $2\pi r = 44$ m.
4. A person walks 60 m in 20 s and then 40 m in 10 s. The average speed for the entire journey is:
Explanation: Total distance = 60 + 40 = 100 m. Total time = 20 + 10 = 30 s. Average speed = 100/30 = 3.33 m/s.
5. A cyclist covers the first half of a journey at 20 km/h and the second half at 30 km/h. The average speed for the whole journey is:
Explanation: For equal distances, average speed = $\frac{2v_1 v_2}{v_1+v_2} = \frac{2\times20\times30}{20+30} = \frac{1200}{50} = 24$ km/h. Note: this is the harmonic mean, not the arithmetic mean.
6. A car moves from A to B (displacement 120 m east) in 15 s. Its average velocity is:
Explanation: Average velocity = displacement / time = 120 m / 15 s = 8 m/s, directed east.
7. A particle moves from x = −5 m to x = 15 m in 4 s. Its average velocity is:
Explanation: Displacement = 15 − (−5) = 20 m. Time = 4 s. Average velocity = 20/4 = 5 m/s (positive x-direction).
8. A particle's position varies as $x = 2t^2 - 3t + 1$ (in metres, t in seconds). Its average velocity between t = 1 s and t = 3 s is:
Explanation: At t=1: x = 2−3+1 = 0 m. At t=3: x = 18−9+1 = 10 m. Average velocity = (10−0)/(3−1) = 10/2 = 5 m/s.
9. If $x = 4t^3 - 2t^2 + t$ (metres), the instantaneous velocity at t = 2 s is:
Explanation: $v = dx/dt = 12t^2 - 4t + 1$. At t=2: $v = 12(4) - 4(2) + 1 = 48 - 8 + 1 = 41$ m/s.
10. For $x = 5\sin(\pi t)$ metres, the instantaneous velocity at t = 0.5 s is:
Explanation: $v = dx/dt = 5\pi\cos(\pi t)$. At t = 0.5 s: $v = 5\pi\cos(\pi/2) = 5\pi\times 0 = 0$ m/s. The particle is momentarily at rest at the peak of its oscillation.
11. A ball is thrown vertically upward. If it rises 20 m before returning to the hand, the total distance travelled and the displacement are:
Explanation: The ball rises 20 m and falls back 20 m, so total distance = 40 m. Since it returns to the starting point, displacement = 0.
12. A particle travels from A to B at 40 m/s and returns from B to A at speed v such that the average speed for the round trip is 48 m/s. The value of v is:
Explanation: Let distance AB = d. Time A→B = d/40, time B→A = d/v. Average speed = 2d/(d/40 + d/v) = 2/(1/40+1/v) = 48. So 1/40+1/v = 2/48 = 1/24. 1/v = 1/24−1/40 = 5/120−3/120 = 2/120 = 1/60. Thus v = 60 m/s.
13. A particle moves along the x-axis. Its position is $x = t^2 - 6t + 5$ m. The particle momentarily stops (instantaneous velocity = 0) at t =
Explanation: $v = dx/dt = 2t - 6$. Setting v = 0: $2t - 6 = 0 \Rightarrow t = 3$ s.
14. Which statement is always true?
Explanation: The path length (distance) is always greater than or equal to the magnitude of displacement. They are equal only when the particle moves in a straight line without reversing direction.
15. Instantaneous velocity is defined as:
Explanation: Instantaneous velocity = $v = \lim_{\Delta t\to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$, the derivative of position with respect to time.
16. A car starts from rest and accelerates at 4 m/s². Its velocity after 5 s is:
Explanation: Using $v = u + at = 0 + 4\times5 = 20$ m/s.
17. A body moving at 10 m/s decelerates uniformly at 2 m/s². The distance it travels before stopping is:
Explanation: Using $v^2 = u^2 + 2as$: $0 = 100 + 2(-2)s \Rightarrow s = 100/4 = 25$ m.
18. A train starts from rest and moves with acceleration 0.5 m/s² for 2 minutes. The distance covered is:
Explanation: t = 2 min = 120 s. $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.5)(120)^2 = 0.25 \times 14400 = 3600$ m.
19. A particle has initial velocity 6 m/s and constant acceleration 2 m/s². The distance covered in the 4th second is:
Explanation: nth second formula: $s_n = u + a(n - \frac{1}{2}) = 6 + 2(4 - 0.5) = 6 + 7 = 13$ m.
20. A body starts from rest with acceleration 3 m/s². The distance covered in the 5th second is:
Explanation: $s_n = u + a(n-\frac{1}{2}) = 0 + 3(5-0.5) = 3\times 4.5 = 13.5$ m.
21. For a body starting from rest with uniform acceleration, the ratio of distances covered in the 1st, 2nd, and 3rd seconds is:
Explanation: $s_n = u + a(n-\frac{1}{2})$. For u=0: $s_1 = a/2$, $s_2 = 3a/2$, $s_3 = 5a/2$. Ratio = 1:3:5. This is a standard JEE result for bodies starting from rest.
22. A stone is dropped from a height of 80 m. Taking g = 10 m/s², the time to reach the ground is:
Explanation: $h = \frac{1}{2}gt^2 \Rightarrow 80 = \frac{1}{2}(10)t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4$ s.
23. A ball is thrown vertically upward at 30 m/s. Taking g = 10 m/s², the maximum height reached is:
Explanation: At maximum height v = 0. $v^2 = u^2 - 2gH \Rightarrow 0 = 900 - 20H \Rightarrow H = 45$ m.
24. A stone is dropped from a tower and reaches the ground in 6 s. Taking g = 10 m/s², the height of the tower is:
Explanation: $h = \frac{1}{2}gt^2 = \frac{1}{2}(10)(36) = 180$ m.
25. A ball is thrown vertically upward from a building roof with speed 20 m/s and falls to the ground 5 s later. Taking g = 10 m/s², the height of the building is:
Explanation: Taking upward as positive, $s = ut - \frac{1}{2}gt^2 = 20(5) - \frac{1}{2}(10)(25) = 100 - 125 = -25$ m. Negative means 25 m below the roof, so the building is 25 m tall.
26. Two trains A and B are approaching each other on parallel tracks. A has velocity 72 km/h and deceleration 1 m/s². B has velocity 54 km/h and deceleration 0.5 m/s². The minimum distance between them initially so they just avoid a collision is:
Explanation: The speeds are 20 m/s and 15 m/s. Their stopping distances are $20^2/(2×1)=200$ m and $15^2/(2×0.5)=225$ m. The minimum initial separation is their sum: 425 m.
27. A particle moves with initial velocity 4 m/s and acceleration −1 m/s². During which numbered second does it reverse direction?
Explanation: The velocity is $v=4-t$, so it becomes zero at $t=4$ s and is negative immediately afterward. Hence the reversal occurs during the 5th second.
28. A particle moving with uniform acceleration travels 20 m in the first 2 s and 60 m in the next 2 s. The acceleration is:
Explanation: In t=0 to 2: $20 = 2u + 2a$ ...(1). In t=0 to 4: $80 = 4u + 8a$ ...(2). From (1): $u + a = 10$. From (2): $u + 2a = 20$. Subtracting: $a = 10$ m/s².
29. A body is released from rest and falls under gravity. The ratio of distances fallen in the 1st second to the 3rd second is:
Explanation: Using $s_n = \frac{1}{2}g(2n-1)$ (distances in successive seconds from rest): $s_1 = g/2$, $s_3 = \frac{1}{2}g(5) = 5g/2$. Ratio = 1:5.
30. A particle starts from rest and has constant acceleration. After 4 s its velocity is 8 m/s. The distance covered in the 4th second alone is:
Explanation: a = 8/4 = 2 m/s², u = 0. $s_4 = u + a(4-0.5) = 0 + 2\times3.5 = 7$ m.
31. Acceleration of a particle is $a = 6t$ m/s². If initial velocity is 2 m/s, the velocity at t = 3 s is:
Explanation: $v = \int a\,dt = \int 6t\,dt = 3t^2 + C$. At t=0, v=2, so C=2. At t=3: $v = 3(9)+2 = 29$ m/s.
32. A particle starts from rest. Its acceleration is $a = 4t$ m/s². The displacement in the first 2 s is:
Explanation: Integrating $a=4t$ gives $v=2t^2$ because the particle starts from rest. Thus $s=\int_0^2 2t^2\,dt=[2t^3/3]_0^2=16/3\text{ m}$.
33. For a particle, $v = 3x^2 + 2$ where x is in metres. The acceleration when x = 2 m is:
Explanation: Use $a=v\,dv/dx$. At $x=2$, $v=3(2)^2+2=14$ m/s and $dv/dx=6x=12\text{ s}^{-1}$. Therefore $a=14×12=168\text{ m/s}^2$.
34. A particle's velocity is $v = (10 - 2t^2)$ m/s. The displacement from t = 0 to t = 2 s is:
Explanation: $s=\int_0^2(10-2t^2)\,dt=[10t-2t^3/3]_0^2=20-16/3=44/3\text{ m}$.
35. The acceleration of a particle is $a = (2t+3)$ m/s², with v = 0 at t = 0. The velocity at t = 2 s is:
Explanation: $v = \int_0^2 (2t+3)dt = [t^2+3t]_0^2 = 4+6 = 10$ m/s.
36. For a particle, $a = v$ (v in m/s, a in m/s²). If v = 1 m/s at t = 0, the velocity at t = 2 s is:
Explanation: $a = dv/dt = v \Rightarrow dv/v = dt \Rightarrow \ln v = t + C$. At t=0, v=1: C=0. So $v = e^t$. At t=2: $v = e^2$ m/s.
37. If $x = t^3 - 6t^2 + 9t + 5$ m, the particle is momentarily at rest at t =
Explanation: $v = dx/dt = 3t^2 - 12t + 9 = 3(t^2-4t+3) = 3(t-1)(t-3)$. $v=0$ at $t=1$ s and $t=3$ s.
38. Acceleration of a particle is $a = -kv^2$ where k is a constant. If initial speed is $v_0$, which expression gives v as a function of time t?
Explanation: $dv/dt = -kv^2 \Rightarrow dv/v^2 = -k\,dt \Rightarrow -1/v = -kt + C$. At t=0, v=v₀: $C = -1/v_0$. So $-1/v = -kt - 1/v_0 \Rightarrow 1/v = 1/v_0 + kt \Rightarrow v = \frac{v_0}{1+kv_0 t}$.
39. The position of a particle is $x = 2t^2 - 8t$ m. The time at which the particle passes through the origin (x = 0) again after t = 0 is:
Explanation: $x = 2t^2 - 8t = 2t(t-4) = 0 \Rightarrow t = 0$ or $t = 4$ s. The particle passes through origin again at t = 4 s.
40. The slope of a velocity-time graph gives:
Explanation: On a v-t graph, $\text{slope} = \Delta v / \Delta t = $ acceleration. This is the defining relation $a = dv/dt$.
41. If $v = \sqrt{4x}$ m/s where x is in metres, the acceleration at x = 4 m is:
Explanation: $v = 2\sqrt{x}$. $\frac{dv}{dx} = \frac{1}{\sqrt{x}}$. $a = v\frac{dv}{dx} = 2\sqrt{x}\cdot\frac{1}{\sqrt{x}} = 2$ m/s² (constant, independent of x).
42. A particle starts from rest at x = 0. Acceleration is $a = 2x$ m/s². The velocity at x = 3 m is:
Explanation: Use $a = v\,dv/dx$: $v\,dv = 2x\,dx \Rightarrow \frac{v^2}{2} = x^2 + C$. At x=0, v=0: C=0. So $v^2 = 2x^2$. At x=3: $v = \sqrt{18} = 3\sqrt{2}$ m/s.
43. Velocity of a particle is $v = 3t^2 - 12t + 9$ m/s. The acceleration when the particle has zero velocity (first time) is:
Explanation: v = 0: $3t^2-12t+9=0 \Rightarrow t^2-4t+3=0 \Rightarrow (t-1)(t-3)=0$. First time: t=1 s. $a = dv/dt = 6t-12$. At t=1: $a = 6-12 = -6$ m/s².
44. A particle's displacement is $s = t^3/3 - 2t^2 + 3t$ metres. The instant at which acceleration is zero is:
Explanation: $v = ds/dt = t^2 - 4t + 3$. $a = dv/dt = 2t - 4 = 0 \Rightarrow t = 2$ s.
45. Area under an acceleration-time graph represents:
Explanation: $\int a\,dt = \Delta v$ (change in velocity). This follows directly from $a = dv/dt$. Similarly, area under v-t graph = displacement.
46. The slope of a position-time (x-t) graph at any point gives:
Explanation: Slope of the x-t graph = $dx/dt$ = instantaneous velocity. Average velocity = slope of the chord (not the tangent).
47. In a v-t graph, the area under the curve between t₁ and t₂ represents:
Explanation: $\int_{t_1}^{t_2} v\,dt = $ displacement (can be negative if v is negative). Distance = $\int|v|dt$, which equals the sum of magnitudes of areas above and below the time axis.
48. A v-t graph is a straight line from (0, 10) to (5, 0). The acceleration is:
Explanation: Slope = $\Delta v / \Delta t = (0-10)/(5-0) = -10/5 = -2$ m/s².
49. In a v-t graph, a straight horizontal line at v = 6 m/s from t = 0 to t = 5 s represents:
Explanation: A horizontal v-t line means velocity is constant — no change in velocity — so acceleration = 0. Displacement = area = 6×5 = 30 m.
50. A v-t graph shows a straight line from (0, 0) to (4, 20). The displacement in 4 s is:
Explanation: Area under v-t graph = area of triangle = ½ × base × height = ½ × 4 × 20 = 40 m.
51. A particle's v-t graph is a parabola. This means the acceleration:
Explanation: If v is a quadratic function of t (parabola), then $a = dv/dt$ is a linear function of t. A straight-line v-t graph → constant a. Parabolic v-t → linearly varying a.
52. From an a-t graph, the area under the curve between t = 1 s and t = 4 s is 9 m/s. If v(1) = 3 m/s, then v(4) is:
Explanation: Area under a-t graph = change in velocity = 9 m/s. $v(4) = v(1) + 9 = 3 + 9 = 12$ m/s.
53. In a position-time graph, a concave upward parabola (opening upward) indicates:
Explanation: If $x = ut + \frac{1}{2}at^2$ with a > 0, the graph is a parabola opening upward (concave up). The slope (velocity) increases with time — positive acceleration.
54. A v-t graph consists of a trapezoid: v = 0 at t = 0, v = 10 m/s from t = 2 s to t = 6 s, and v = 0 at t = 8 s. The total displacement is:
Explanation: Area = triangle (0 to 2) + rectangle (2 to 6) + triangle (6 to 8) = ½×2×10 + 4×10 + ½×2×10 = 10+40+10 = 60 m.
55. On a v-t graph, a line with negative slope crossing the time axis represents:
Explanation: A line starting at positive v and crossing the t-axis means v becomes negative after the crossing. The particle first slows down, momentarily stops (v=0), then moves in the negative direction.
56. A body moves with velocity $v = 4 - t$ m/s. The distance covered from t = 0 to t = 6 s is:
Explanation: v = 0 at t = 4 s. Distance 0 to 4 s = $\int_0^4 (4-t)dt = [4t-t^2/2]_0^4 = 16-8 = 8$ m. Distance 4 to 6 s = $|\int_4^6 (4-t)dt| = |[4t-t^2/2]_4^6| = |(24-18)-(16-8)| = |6-8| = 2$ m. Total distance = 8+2 = 10 m.
57. The x-t graph of a particle is a straight line passing through the origin with slope 5 m/s. The v-t graph of this particle is:
Explanation: A straight-line x-t graph means x = 5t, so v = dx/dt = 5 m/s = constant. On a v-t graph, constant velocity is a horizontal line.
58. For a particle with $x = 3 + 5t - t^2$ m, the particle returns to its initial position at t =
Explanation: Initial position x(0) = 3 m. $3 + 5t - t^2 = 3 \Rightarrow 5t - t^2 = 0 \Rightarrow t(5-t) = 0$. So t = 0 or t = 5 s. The particle returns at t = 5 s.
59. In a v-t graph for uniformly accelerated motion, which is correct?
Explanation: For $v = u + at$ (uniform acceleration a = constant), v is a linear function of t. The v-t graph is a straight line. Slope = a (acceleration). Area = displacement.
60. A particle has v = 10 m/s at t = 0. From t = 0 to t = 3 s, the a-t graph is a horizontal line at a = 4 m/s². From t = 3 s to t = 5 s, a = −6 m/s². The velocity at t = 5 s is:
Explanation: $\Delta v$ (0 to 3) = 4×3 = 12 m/s. v(3) = 10+12 = 22 m/s. $\Delta v$ (3 to 5) = −6×2 = −12 m/s. v(5) = 22−12 = 10 m/s.