Kinematics in a straight line: AS Level Mathematics questions with answers
Free Kinematics in a straight line questions for AS 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 20 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)
0.857 s, 13.5 m
sA = 20t − 5t2B1 sB = 15t + 5t2B1 sA + sB = 30 M1 t = 0.857 A1 height 13.5 A1
5
(b)
A 11.4 m s−1 upwards, B 23.6 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 − 8t + 15 for 0 ≤ t ≤ 6.
(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 = 6.
[2]
(c)
Find the total distance travelled by P in the 6 seconds.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
−2 m s−2 at t = 3; 2 m s−2 at t = 5
v = 0 solved M1 a = dvdt = 2t − 8 M1 both values A1
3
(b)
18 m
integrate v M1 18 A1
2
(c)
623 m
split the interval at the times of rest M1 distances 18, 43, 43A1 sum of magnitudes M1 623A1
4
Question 3Structured · 4 marks
A car accelerates uniformly from 2 m s−1 to 21 m s−1 in 7 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) 2.71 m s−2 (b) 80.5 m
v = u + at M1 2.71 A1 s = ½(u + v)t M1 80.5 A1
4
Question 4Structured · 5 marks
A particle starts from rest at O and moves in a straight line with acceleration a = (8 − 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)
16 m s−1
integrate a M1 v = 8t − t2A1 t = 4: v = 16 A1
3
(b)
1283 m
integrate v M1 1283A1
2
Question 5Structured · 6 marks
A particle moves in a straight line. Its velocity t seconds after leaving O is v = t(6 − t) m s−1 for 0 ≤ t ≤ 6.
(a)
Find the maximum velocity.
[3]
(b)
Find the distance travelled in the 6 seconds.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
9 m s−1
dvdt = 0 M1 t = 3 A1 9 A1
3
(b)
36 m
integrate M1 limits 0 to 6 M1 36 A1
3
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 10 m s−1 from a point 20 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)
0.571 s, 12.7 m
sA = 25t − 5t2B1 sB = 10t + 5t2B1 sA + sB = 20 M1 t = 0.571 A1 height 12.7 A1
5
(b)
A 19.3 m s−1 upwards, B 15.7 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