Kinematics in a straight line: A Level Mathematics questions with answers
Free Kinematics in a straight line questions for A Level Mathematics (9709), part of Mechanics. Each question has the answer and the marking points an examiner would look for.
Particle A is projected vertically upwards from ground level with speed 30 m s−1. At the same instant particle B is projected vertically downwards with speed 15 m s−1 from a point 30 m directly above A. Find the time at which the particles meet and their height above the ground.
[5]
(b)
Find the velocity of each particle at the instant they meet.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.667 s, 17.8 m
sA = 30t − 5t2B1 sB = 15t + 5t2B1 sA + sB = 30 M1 t = 0.667 A1 height 17.8 A1
5
(b)
A 23.3 m s−1 upwards, B 21.7 m s−1 downwards
v = u + at for each M1 both correct with directions A1
2
Question 2Structured · 9 marksA* challenge
A particle P moves in a straight line, starting from rest at O. Its velocity at time t seconds after leaving O is v m s−1, where v = t2 − 4t + 3 for 0 ≤ t ≤ 4.
(a)
Find the acceleration of P at each instant when it is at instantaneous rest.
[3]
(b)
Find the displacement of P from O at t = 4.
[2]
(c)
Find the total distance travelled by P in the 4 seconds.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
−2 m s−2 at t = 1; 2 m s−2 at t = 3
v = 0 solved M1 a = dvdt = 2t − 4 M1 both values A1
3
(b)
43 m
integrate v M1 43A1
2
(c)
4 m
split the interval at the times of rest M1 distances 43, 43, 43A1 sum of magnitudes M1 4 A1
4
Question 3Structured · 4 marks
A car accelerates uniformly from 6 m s−1 to 17 m s−1 in 10 s. Find (a) the acceleration, (b) the distance travelled in this time.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a) 1.1 m s−2 (b) 115 m
v = u + at M1 1.1 A1 s = ½(u + v)t M1 115 A1
4
Question 4Structured · 6 marks
A particle moves in a straight line. Its velocity t seconds after leaving O is v = t(4 − t) m s−1 for 0 ≤ t ≤ 4.
(a)
Find the maximum velocity.
[3]
(b)
Find the distance travelled in the 4 seconds.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
4 m s−1
dvdt = 0 M1 t = 2 A1 4 A1
3
(b)
323 m
integrate M1 limits 0 to 4 M1 323A1
3
Question 5Structured · 5 marks
A particle starts from rest at O and moves in a straight line with acceleration a = (4 − 2t) m s−2.
(a)
Find the velocity when the acceleration is zero.
[3]
(b)
Find the displacement from O at this time.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
4 m s−1
integrate a M1 v = 4t − t2A1 t = 2: v = 4 A1
3
(b)
163 m
integrate v M1 163A1
2
Question 6Structured · 7 marksA* challenge
(a)
Particle A is projected vertically upwards from ground level with speed 25 m s−1. At the same instant particle B is projected vertically downwards with speed 15 m s−1 from a point 40 m directly above A. Find the time at which the particles meet and their height above the ground.
[5]
(b)
Find the velocity of each particle at the instant they meet.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
1 s, 20 m
sA = 25t − 5t2B1 sB = 15t + 5t2B1 sA + sB = 40 M1 t = 1 A1 height 20 A1
5
(b)
A 15 m s−1 upwards, B 25 m s−1 downwards
v = u + at for each M1 both correct with directions A1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in