IMO Practice Test — Motion in a Straight Line
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
A particle's position is x = 2t^3 - 9t^2 + 12t (SI). At what time is its velocity zero (smallest positive t)?
1 s
1.5 s
2 s
3 s
Explanation: $v = 6t^2 - 18t + 12 = 6(t-1)(t-2)$; zero at $t = 1\,\text{s}$ and $t = 2\,\text{s}$, smallest is $1\,\text{s}$.
Question 2 of 14
A car covers the first half of a distance at 20 m/s and the second half at 30 m/s. Its average speed is:
24 m/s
25 m/s
26 m/s
12 m/s
Explanation: Equal distances: $v_{avg} = \dfrac{2 v_1 v_2}{v_1 + v_2} = \dfrac{2 \times 20 \times 30}{50} = 24\,\text{m/s}$.
Question 3 of 14
A ball thrown up returns to the thrower in 6 s. The initial speed was (g = 10 m/s$^2$):
15 m/s
30 m/s
45 m/s
60 m/s
Explanation: Time of flight $T = \dfrac{2u}{g} \Rightarrow u = \dfrac{gT}{2} = \dfrac{10 \times 6}{2} = 30\,\text{m/s}$.
Question 4 of 14
A body starts from rest with uniform acceleration. The ratio of distances covered in the 1st, 2nd and 3rd seconds is:
1 : 2 : 3
1 : 3 : 5
1 : 4 : 9
2 : 3 : 4
Explanation: Distances in successive seconds go as $(2n - 1)$: $1 : 3 : 5$.
Question 5 of 14
A stone is dropped from a balloon ascending at 10 m/s when at height 60 m. Time to reach the ground is (g = 10 m/s$^2$):
2 s
4 s
6 s
8 s
Explanation: Taking down positive with initial velocity $-10$: $60 = -10t + 5t^2 \Rightarrow t^2 - 2t - 12 = 0 \Rightarrow t = 1 + \sqrt{13} \approx 4\,\text{s}$.
Question 6 of 14
Two trains 120 m and 80 m long move toward each other at 15 m/s and 10 m/s. Time to completely cross each other is:
6 s
8 s
10 s
12 s
Explanation: Relative speed $= 25\,\text{m/s}$; total length $= 200\,\text{m}$; $t = \dfrac{200}{25} = 8\,\text{s}$.
Question 7 of 14
The displacement of a body is zero but the distance is not. The body has:
Constant velocity
Returned to its start
Zero acceleration
Infinite speed
Explanation: Zero displacement with non-zero path length means it came back to the starting point.
Question 8 of 14
A v-t graph is a straight line passing through the origin with positive slope. The motion is:
Uniform velocity
Uniformly accelerated from rest
Retarded
At rest
Explanation: Through the origin means $u = 0$; constant positive slope means constant acceleration.
Question 9 of 14
A car decelerates uniformly from 40 m/s and stops in 100 m. Its deceleration is:
4 m/s$^2$
8 m/s$^2$
10 m/s$^2$
16 m/s$^2$
Explanation: $0 = 40^2 + 2a(100) \Rightarrow a = -\dfrac{1600}{200} = -8\,\text{m/s}^2$.
Question 10 of 14
A particle moving with uniform acceleration covers 40 m in the 4th second and 60 m in the 6th second. Its acceleration is:
5 m/s$^2$
10 m/s$^2$
15 m/s$^2$
20 m/s$^2$
Explanation: $s_n = u + \dfrac{a}{2}(2n-1)$; subtracting: $60 - 40 = \dfrac{a}{2}(11 - 7) = 2a \Rightarrow a = 10\,\text{m/s}^2$.
Question 11 of 14
From the top of a tower a stone is thrown up at 20 m/s and another dropped at the same instant. The difference in their speeds after 2 s is (g = 10 m/s$^2$):
0 m/s
10 m/s
20 m/s
40 m/s
Explanation: Both gain the same $gt$, so the difference in velocities stays equal to the initial difference, $20\,\text{m/s}$.
Question 12 of 14
A body travels half its total path in the last second of free fall from rest. The total time of fall is about:
2.4 s
3.4 s
4.4 s
5.4 s
Explanation: If total time is $t$, last-second fraction $\dfrac{1}{2}$ gives $(t-1)^2 = \dfrac{t^2}{2}$, so $t = 2 + \sqrt{2} \approx 3.4\,\text{s}$.
Question 13 of 14
On an x-t graph, a curve bending so its slope increases with time indicates:
Negative acceleration
Zero acceleration
Positive acceleration
Constant velocity
Explanation: Increasing slope of $x$-$t$ means increasing velocity, i.e. positive acceleration.
Question 14 of 14
A man rows a boat at 5 m/s in still water across a river flowing at 3 m/s. Relative to the bank, considering only the downstream direction, the river carries the boat downstream at:
2 m/s
3 m/s
4 m/s
8 m/s
Explanation: The downstream drift is set by the river's velocity along the bank, $3\,\text{m/s}$ (the rowing is across the flow).