Convert a time of 4 minutes 30 seconds into seconds.
(Total for Question 3 is 1 mark)
4
A particle moves at a constant speed of 15 m/s for 6 seconds. Find the distance it travels.
(Total for Question 4 is 1 mark)
5
A particle has initial speed 3 m/s and accelerates uniformly at 2 m/s2 for 9 seconds. Find its final speed.
(Total for Question 5 is 2 marks)
6
A cyclist accelerates uniformly from 2 m/s to 14 m/s in 4 seconds. Find the acceleration.
(Total for Question 6 is 2 marks)
7
A car decelerates uniformly from 24 m/s to rest in 6 seconds. Find the distance travelled while decelerating.
(Total for Question 7 is 2 marks)
8
A particle moves with initial speed 2 m/s and constant acceleration 3 m/s2. Find the distance it travels in the first 4 seconds.
(Total for Question 8 is 2 marks)
9
A particle accelerates uniformly from 5 m/s to 13 m/s while travelling 36 m. Find the acceleration.
(Total for Question 9 is 2 marks)
10
A ball is released from rest and falls freely under gravity, g = 9.8 m/s2. Find its speed after falling for 1 second.
(Total for Question 10 is 1 mark)
11
A stone is dropped from rest and takes 2 seconds to reach the ground. Using g = 9.8 m/s2, find the height fallen.
(Total for Question 11 is 2 marks)
12
A particle P has position vector (2i - 5j) m at time t = 0 and moves with constant velocity (4i + 3j) m/s. Find the position vector of P at time t = 3 seconds.
(Total for Question 12 is 2 marks)
13
A particle moves with velocity v = (5i - 12j) m/s. Find the speed of the particle.
(Total for Question 13 is 2 marks)
14
The displacement of a particle at time t seconds is x = 4t2 - 3t metres. Find an expression for the velocity v m/s at time t.
(Total for Question 14 is 1 mark)
15
A particle moves in a straight line with constant acceleration. It passes point A with speed 4 m/s. Nine seconds later it passes point B with speed 22 m/s. Find the distance AB.
(Total for Question 15 is 3 marks)
16
A ball is thrown vertically downwards from the top of a cliff with initial speed 6 m/s. It hits the sea 2.5 seconds later. Using g = 9.8 m/s2, find the height of the cliff, giving your answer to 3 significant figures.
(Total for Question 16 is 3 marks)
17
A train travels along a straight track. It accelerates uniformly from rest to a speed of 20 m/s in 25 seconds, then travels at this constant speed for a further 40 seconds, before decelerating uniformly to rest in 10 seconds. Find the total distance travelled by the train.
(Total for Question 17 is 4 marks)
18
A particle P moves along the x-axis. At time t seconds (t ≥ 0) its displacement from a fixed point O is x metres, where x = t3 - 9t2 + 24t. Find the two values of t at which P is instantaneously at rest, and find the acceleration of P at the larger of these two values of t.
(Total for Question 18 is 4 marks)
19
At time t = 0, a cyclist passes a fixed point O with constant speed 6 m/s, travelling in a straight line. At the same instant, a runner sets off from rest from O, moving in the same direction with constant acceleration 1.5 m/s2. Find the time at which the runner catches up with the cyclist, and the distance from O at which this happens.
(Total for Question 19 is 4 marks)
Mark scheme · M1D Mechanics: Quantities, Units and Kinematics: Fluency and Exam Drill
Question 1
B1 m/s2 (or m s-2) oe
Answer: m/s2
Question 2
B1 5600 cao
Answer: 5600 m
Question 3
B1 270 cao
Answer: 270 s
Question 4
B1 90 cao
Answer: 90 m
Question 5
M1 use v = u + at with u=3, a=2, t=9
A1 21 m/s cao
Answer: 21 m/s
Question 6
M1 a = (14-2)/4
A1 3 m/s2 cao
Answer: 3 m/s2
Question 7
M1 s = 1/2(u+v)t with u=24, v=0, t=6
A1 72 m cao
Answer: 72 m
Question 8
M1 s = ut + 1/2 a t2 with u=2, a=3, t=4
A1 32 m cao
Answer: 32 m
Question 9
M1 use v2 = u2 + 2as with u=5, v=13, s=36
A1 2 m/s2 cao
Answer: 2 m/s2
Question 10
B1 9.8 cao
Answer: 9.8 m/s
Question 11
M1 s = ut + 1/2 a t2 with u=0, a=9.8, t=2
A1 19.6 m cao
Answer: 19.6 m
Question 12
M1 r = r0 + vt applied with t=3
A1 (14i + 4j) m cao
Answer: (14i + 4j) m
Question 13
M1 speed = √52 + 122
A1 13 m/s cao
Answer: 13 m/s
Question 14
B1 v = 8t - 3 cao
Answer: v = 8t - 3
Question 15
M1 a = (22-4)/9
M1 s = 1/2(u+v)t = 1/2(4+22)(9), or ft using s = ut + 1/2 a t2
A1 117 m cao
Answer: 117 m
Question 16
M1 s = ut + 1/2 a t2 with u=6, a=9.8, t=2.5
M1 s = 6*2.5 + 0.5*9.8*2.52 correctly evaluated
A1 awrt 45.6 m
Answer: 45.6 m (3 s.f.)
Question 17
M1 distance in first stage = 1/2(0+20)(25) = 250
M1 distance at constant speed = 20*40 = 800
M1 distance decelerating = 1/2(20+0)(10) = 100
A1 1150 m cao (250+800+100)
Answer: 1150 m
Question 18
M1 differentiate x to find v = 3t2 - 18t + 24
M1 set v=0 and solve, e.g. factorise 3(t-2)(t-4)=0
A1 t = 2 and t = 4 cao
A1 a = 6t - 18, so a(4) = 6 m/s2 cao
Answer: t = 2 s and t = 4 s; acceleration at t = 4 is 6 m/s2
Question 19
M1 runner's distance from O is 1/2(1.5)t2 = 0.75t2, using s = ut + 1/2 a t2 with u=0