IMO Practice Test — Motion
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
A particle goes 3 m east, 4 m north and 3 m west. The magnitude of displacement is:
10 m
4 m
6 m
0 m
Explanation: Net east-west = 3 - 3 = 0; net north = 4 m, so displacement = 4 m.
Question 2 of 14
A body covers half its journey at 30 km/h and the other half at 60 km/h. Its average speed is:
45 km/h
40 km/h
50 km/h
90 km/h
Explanation: For equal distances, avg speed = 2(30)(60)/(30+60) = 3600/90 = 40 km/h.
Question 3 of 14
A car starting from rest covers 20 m in the first 2 s of uniform acceleration. The distance in the next 2 s is:
20 m
40 m
60 m
80 m
Explanation: a = 10 m/s$^2$ (since 20 = (1/2)a(4)). In 4 s, s = (1/2)(10)(16) = 80 m; next 2 s = 80 - 20 = 60 m.
Question 4 of 14
On a velocity-time graph, the line goes from (0, 0) to (4 s, 8 m/s) then stays flat to 10 s. Total displacement is:
64 m
32 m
48 m
56 m
Explanation: Triangle: (1/2)(4)(8) = 16 m; rectangle: 8 x 6 = 48 m; total = 16 + 48 = 64 m.
Question 5 of 14
Two trains move towards each other at 15 m/s and 25 m/s, starting 800 m apart. They meet after:
20 s
32 s
40 s
10 s
Explanation: Relative speed = 15 + 25 = 40 m/s; time = 800/40 = 20 s.
Question 6 of 14
A ball thrown up returns to the thrower's hand. Over the whole flight, its displacement and distance are:
Both zero
Displacement zero, distance non-zero
Both equal
Distance zero, displacement non-zero
Explanation: It returns to start, so displacement = 0, but it travelled up and down, so distance is non-zero.
Question 7 of 14
An object moves so that s = 5t (s in m, t in s). Its acceleration is:
5 m/s$^2$
0 m/s$^2$
10 m/s$^2$
2.5 m/s$^2$
Explanation: s is proportional to t means constant velocity (5 m/s), so acceleration = 0.
Question 8 of 14
A car decelerates uniformly from 30 m/s and stops in 6 s. The distance covered is:
90 m
180 m
45 m
30 m
Explanation: Use s = (u+v)/2 x t = (30+0)/2 x 6 = 15 x 6 = 90 m.
Question 9 of 14
In uniform acceleration from rest, the ratio of distances covered in the 1st, 2nd and 3rd seconds is:
1 : 2 : 3
1 : 3 : 5
1 : 4 : 9
1 : 1 : 1
Explanation: Distance in nth second is proportional to (2n - 1), giving 1 : 3 : 5.
Question 10 of 14
A body has u = 10 m/s and a = -2 m/s$^2$. The time to reach maximum displacement (momentarily at rest) is:
2 s
5 s
10 s
20 s
Explanation: v = u + at => 0 = 10 - 2t => t = 5 s.
Question 11 of 14
The displacement-time graph of a body is a parabola opening upward. This indicates:
Constant velocity
Uniform positive acceleration
Object at rest
Retardation only
Explanation: s proportional to t^2 (a parabola) corresponds to uniform positive acceleration.
Question 12 of 14
A car at 72 km/h brakes at 5 m/s$^2$. Its stopping distance is:
20 m
40 m
80 m
10 m
Explanation: 72 km/h = 20 m/s; 0 = 20^2 - 2(5)s => s = 400/10 = 40 m.
Question 13 of 14
Two bodies start together; A has constant velocity 10 m/s, B starts from rest at a = 2 m/s$^2$. B catches A after:
5 s
10 s
20 s
2 s
Explanation: 10t = (1/2)(2)t^2 => 10t = t^2 => t = 10 s.
Question 14 of 14
The area between a v-t line and the time axis below the axis (negative velocity) represents:
Positive displacement
Displacement in the opposite direction
Distance only
Acceleration
Explanation: Area below the time axis is negative, i.e. displacement in the opposite direction.