O Level Additional Mathematics (4037)
Kinematics
7 questions · about 4% of the marks we indexed · examined in 7 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Displacement, velocity, acceleration · Core · Paper 1 style · 5 marksA particle moves in a straight line so that, t seconds after passing through a fixed point O, its displacement from O is s m, where s = t³ − 9t² + 24t.
- (a) Find the values of t when the particle is instantaneously at rest. [3]
- (b) Find the acceleration of the particle when t = 5. [2]
Show answer
- (a) t = 2 and t = 4
v = 3t² − 18t + 24 B1; = 3(t − 2)(t − 4) = 0 M1; t = 2, 4 A1
- (b) 12 m/s²
a = 6t − 18 M1; = 12 A1
A particle starts from a fixed point O and moves in a straight line. Its velocity t seconds later is v m/s, where v = 5t − t².
- (a) Find the acceleration of the particle when t = 1. [2]
- (b) Find the displacement of the particle from O when t = 3. [3]
Show answer
- (a) 3 m/s²
a = 5 − 2t M1 = 3 A1
- (b) 13.5 m
s = 5t²/2 − t³/3 (+ c, c = 0) M1 A1; s(3) = 22.5 − 9 = 13.5 A1
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity is v m/s, where v = 3t² − 2t + 1.
- (a) Find the acceleration of the particle when t = 2. [2]
- (b) Find the displacement of the particle from O when t = 2. [3]
Show answer
- (a) 10 m/s²
a = 6t − 2 M1 = 10 A1
- (b) 6 m
s = t³ − t² + t (c = 0) M1 A1; 8 − 4 + 2 = 6 A1
A cyclist sets off from rest along a straight road. Her velocity t seconds later is v m/s, where v = 0.3t(10 − t) for 0 ≤ t ≤ 10.
- (a) Find the maximum speed of the cyclist. [2]
- (b) Find the distance travelled by the cyclist in the 10 seconds. [3]
Show answer
- (a) 7.5 m/s
dv/dt = 3 − 0.6t = 0, t = 5 M1; v = 7.5 A1
- (b) 50 m
∫ (3t − 0.3t²) dt = 1.5t² − 0.1t³ M1 A1; 150 − 100 = 50 A1
A particle passes through a fixed point O with velocity 3 m/s and moves in a straight line. Its acceleration t seconds later is (2t + 1) m/s².
- (a) Find the velocity of the particle when t = 3. [2]
- (b) Find the displacement of the particle from O when t = 3. [2]
Show answer
- (a) 15 m/s
v = t² + t + 3 M1; 9 + 3 + 3 = 15 A1
- (b) 22.5 m
s = t³/3 + t²/2 + 3t M1; 9 + 4.5 + 9 = 22.5 A1
A particle moves in a straight line. Its velocity t seconds after leaving a fixed point is v m/s, where v = 6e^(−0.5t).
- (a) Write down the initial velocity of the particle. [1]
- (b) Find the acceleration when t = 2, correct to 3 significant figures. [2]
- (c) Find the distance travelled in the first 4 seconds, correct to 3 significant figures. [3]
Show answer
- (a) 6 m/s
B1
- (b) −1.10 m/s²
a = −3e^(−0.5t) M1; −3e⁻¹ = −1.10 A1
- (c) 10.4 m
s = −12e^(−0.5t) + c M1; [−12e^(−0.5t)] from 0 to 4 M1; 12(1 − e⁻²) = 10.4 A1 (v > 0 throughout)
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its displacement from O is s m, where s = 3 sin 2t.
- (a) Find an expression for the velocity of the particle. [1]
- (b) Find the first time after t = 0 at which the particle is at rest, and its acceleration at that time. [3]
Show answer
- (a) v = 6 cos 2t
B1
- (b) t = π/4; a = −12 m/s²
cos 2t = 0, t = π/4 M1 A1; a = −12 sin 2t = −12 A1
A particle moves in a straight line with acceleration a m/s², where a = 6 − 2t, t seconds after it passes a point P. Its velocity at P is 4 m/s.
- (a) Find an expression for v in terms of t. [2]
- (b) Find the maximum velocity of the particle and the time at which it occurs. [3]
Show answer
- (a) v = 4 + 6t − t²
integrate a M1; c = 4 A1
- (b) 13 m/s at t = 3
a = 0 when t = 3 M1 A1; v = 4 + 18 − 9 = 13 A1
A particle moves in a straight line so that its velocity t seconds after leaving a point O is v m/s, where v = 10 − 8e^(−t/4).
- (a) Find the initial velocity and acceleration of the particle. [2]
- (b) Find the time at which the velocity is 8 m/s, correct to 3 significant figures, and the acceleration at that time. [4]
- (c) State the value that the velocity approaches as t becomes very large. [1]
Show answer
- (a) v = 2 m/s, a = 2 m/s²
B1 B1 (a = 2e^(−t/4))
- (b) t = 5.55 s; a = 0.5 m/s²
e^(−t/4) = 0.25 M1; t = 4 ln 4 = 5.55 A1; a = 2 × 0.25 M1 = 0.5 A1
- (c) 10 m/s
B1
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its displacement from O is s m, where s = ln(t + 1) − t/3.
- (a) Find the time at which the particle is at rest. [2]
- (b) Find the exact displacement of the particle from O at this time. [2]
- (c) Find the acceleration of the particle at this time. [2]
Show answer
- (a) t = 2
v = 1/(t + 1) − ⅓ M1; t = 2 A1
- (b) ln 3 − ⅔ m
s(2) = ln 3 − 2/3 B1 B1
- (c) −1/9 m/s²
a = −1/(t + 1)² M1 = −1/9 A1
A lift in an office tower in Dubai rises from rest. Its velocity t seconds later is v m/s, where v = 0.6t² − 0.1t³, for 0 ≤ t ≤ 6.
- (a) Show that the lift is at rest when t = 6. [1]
- (b) Find the maximum speed of the lift. [3]
- (c) Find the height through which the lift rises in the 6 seconds. [2]
Show answer
- (a) v(6) = 21.6 − 21.6 = 0
B1
- (b) 3.2 m/s
dv/dt = 1.2t − 0.3t² = 0 M1; t = 4 A1; v = 3.2 A1
- (c) 10.8 m
∫₀⁶ v dt = [0.2t³ − 0.025t⁴] M1 = 10.8 A1
A particle moves in a straight line so that, t seconds after observations begin, its displacement from a fixed point O is s m, where s = 2t³ − 15t² + 24t + 5.
- (a) Find the values of t when the particle is at rest and its displacements from O at these times. [4]
- (b) Find the total distance travelled in the first 4 seconds. [2]
Show answer
- (a) t = 1, s = 16; t = 4, s = −11
v = 6t² − 30t + 24 = 6(t − 1)(t − 4) M1 A1; s(1) = 16, s(4) = −11 A1 A1
- (b) 38 m
s(0) = 5: 11 + 27 M1 = 38 A1
A particle starts from rest at O and moves in a straight line with velocity v = kt(8 − t) m/s, where k is a constant. It travels 36 m in the first 6 seconds.
- (a) Find the value of k. [3]
- (b) Find the acceleration of the particle when t = 6. [2]
Show answer
- (a) k = 0.5
∫₀⁶ k(8t − t²) dt = k[4t² − t³/3] M1 A1; 72k = 36, k = 0.5 A1
- (b) −2 m/s²
a = k(8 − 2t) M1 = −2 A1
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity is v m/s, where v = t² − 5t + 4.
- (a) Find the times when the particle is instantaneously at rest. [2]
- (b) Find the total distance travelled by the particle in the first 5 seconds. [5]
Show answer
- (a) t = 1 and t = 4
(t − 1)(t − 4) = 0 M1 A1
- (b) 49/6 m
s = t³/3 − 5t²/2 + 4t M1 A1; s(1) = 11/6, s(4) = −8/3, s(5) = −5/6 M1 A1; 11/6 + 27/6 + 11/6 = 49/6 A1
Two particles A and B leave a fixed point O at the same time and move in the same straight line. After t seconds, A has velocity (4t + 1) m/s and B has velocity 3t² m/s.
- (a) Find the exact value of t when A and B next meet. [4]
- (b) Find the velocities of A and B at this time, in the form a + b√2. [2]
Show answer
- (a) t = 1 + √2
s_A = 2t² + t, s_B = t³ B1; t³ − 2t² − t = 0 M1; t² − 2t − 1 = 0 M1; t = 1 + √2 A1
- (b) v_A = 5 + 4√2, v_B = 9 + 6√2
B1 B1
A particle starts from rest at a fixed point O and moves in a straight line so that its velocity after t seconds is v m/s, where v = 2 sin 3t.
- (a) Find an expression for the displacement from O at time t. [2]
- (b) Find the greatest distance of the particle from O. [1]
- (c) Find the total distance travelled by the particle for 0 ≤ t ≤ π. [3]
Show answer
- (a) s = ⅔(1 − cos 3t)
−⅔ cos 3t + c M1; c = ⅔ A1
- (b) 4/3 m
B1
- (c) 4 m
rest at t = π/3, 2π/3 M1; s = 4/3, 0, 4/3 at π/3, 2π/3, π M1; 3 × 4/3 = 4 A1
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity is v m/s, where v = e^(2t) − 4eᵗ + 3.
- (a) Find the exact value of t, other than t = 0, at which the particle is at rest, and its acceleration at that time. [4]
- (b) Find the exact distance travelled by the particle between t = 0 and this time. [3]
Show answer
- (a) t = ln 3; a = 6 m/s²
(eᵗ − 1)(eᵗ − 3) = 0 M1; t = ln 3 A1; a = 2e^(2t) − 4eᵗ M1 = 18 − 12 = 6 A1
- (b) 4 − 3 ln 3 m
s = ½e^(2t) − 4eᵗ + 3t M1; [ ] from 0 to ln 3 = 3 ln 3 − 4 A1; distance 4 − 3 ln 3 A1
A particle passes through O with velocity 6 m/s. Its acceleration t seconds later is a m/s², where a = −4/(t + 1)².
- (a) Find an expression for v in terms of t, and explain why the particle never comes to rest. [3]
- (b) Find the displacement of the particle from O when t = 3, correct to 3 significant figures. [3]
Show answer
- (a) v = 4/(t + 1) + 2; v > 2 for all t ≥ 0
integrate M1; c = 2 A1; 4/(t + 1) > 0 so v > 2 B1
- (b) 11.5 m
s = 4 ln(t + 1) + 2t (c = 0) M1 A1; 4 ln 4 + 6 = 11.5 A1
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its displacement from O is s m, where s = te^(−t/2).
- (a) Find the time at which the particle is instantaneously at rest and its greatest distance from O, giving the distance correct to 3 significant figures. [4]
- (b) Find the acceleration at this time, correct to 3 significant figures. [3]
Show answer
- (a) t = 2; 0.736 m
v = e^(−t/2)(1 − t/2) M1 A1; t = 2 A1; s = 2e⁻¹ = 0.736 A1
- (b) −0.184 m/s²
a = e^(−t/2)(t/4 − 1) M1 A1; −½e⁻¹ = −0.184 A1
A car on the M2 motorway starts from rest. Its velocity-time graph consists of three straight-line segments joining the points (0, 0), (10, 15), (40, 15) and (50, 0), where t is in seconds and v in m/s.
- (a) Find the acceleration in the first 10 seconds. [1]
- (b) Find the total distance travelled. [3]
Show answer
- (a) 1.5 m/s²
15 ÷ 10 B1
- (b) 600 m
area under graph M1; 75 + 450 + 75 M1 = 600 A1
A bus starts from rest and accelerates uniformly at 0.8 m/s² until it reaches a speed of 12 m/s. It travels at this speed for 60 seconds and then decelerates uniformly at 1.2 m/s² until it stops.
- (a) Find the time taken to reach 12 m/s. [1]
- (b) Find the time taken to come to rest. [1]
- (c) Find the total distance travelled by the bus. [3]
Show answer
- (a) 15 s
12 ÷ 0.8 B1
- (b) 10 s
12 ÷ 1.2 B1
- (c) 870 m
90 + 720 + 60 M1 A1 A1
The displacement-time graph of a runner doing shuttle drills consists of straight lines joining (0, 0), (4, 20), (6, 20) and (10, 0), where t is in seconds and s is in metres.
- (a) Find the velocity of the runner in each of the three stages. [3]
- (b) Find the average speed of the runner for the whole 10 seconds. [2]
Show answer
- (a) 5 m/s, 0 m/s, −5 m/s
B1 B1 B1
- (b) 4 m/s
total distance 40 m M1; 40 ÷ 10 = 4 A1
A particle moves in a straight line so that its velocity t seconds after passing O is v m/s, where v = 12 − 3t.
- (a) Sketch the velocity-time graph for 0 ≤ t ≤ 6, stating where it crosses each axis. [2]
- (b) Find the displacement of the particle from O when t = 6. [2]
- (c) Find the total distance travelled in the 6 seconds. [2]
Show answer
- (a) straight line from (0, 12) crossing the t-axis at t = 4, ending at (6, −6)
B1 line; B1 intercepts
- (b) 18 m
s = 12t − 1.5t² M1; 72 − 54 = 18 A1
- (c) 30 m
24 + 6 M1 A1
A train leaves a station from rest and accelerates uniformly to a speed of V m/s in 40 s. It travels at V m/s for 3 minutes and then decelerates uniformly to rest in 60 s. The distance between the stations is 4.2 km.
- (a) Sketch the velocity-time graph for the journey. [1]
- (b) Find the value of V, correct to 3 significant figures. [3]
- (c) Find the deceleration of the train. [1]
Show answer
- (a) trapezium: rising line for 40 s, horizontal for 180 s, falling line for 60 s
B1
- (b) 18.3 m/s
area = ½(40)V + 180V + ½(60)V = 230V M1 A1; V = 4200/230 = 18.3 A1
- (c) 0.304 m/s²
18.26 ÷ 60 B1
A particle moves in a straight line so that its velocity t seconds after leaving a point O is v m/s, where v = 4t − t².
- (a) Sketch the velocity-time graph for 0 ≤ t ≤ 5, stating the values of t where it meets the t-axis and the maximum velocity. [3]
- (b) Find the total distance travelled for 0 ≤ t ≤ 5. [4]
Show answer
- (a) inverted parabola through (0, 0) and (4, 0), maximum (2, 4), v(5) = −5
shape B1; intercepts 0 and 4 B1; max (2, 4) and end (5, −5) B1
- (b) 13 m
s = 2t² − t³/3 M1; 0 to 4: 32/3 A1; 4 to 5: −7/3 A1; total 13 A1
The velocity-time graph of a particle consists of straight lines joining the points (0, 2), (4, 10), (10, 10) and (14, 0), where t is in seconds and v is in m/s.
- (a) Find the acceleration during the first 4 seconds. [1]
- (b) Find the total distance travelled. [3]
- (c) Find the time at which the particle has travelled half of the total distance. [2]
Show answer
- (a) 2 m/s²
(10 − 2) ÷ 4 B1
- (b) 104 m
24 + 60 + 20 M1 A1 A1
- (c) t = 6.8 s
need 52 − 24 = 28 m at 10 m/s M1; t = 4 + 2.8 A1
A particle starts from rest. Its acceleration is 3 m/s² for 0 ≤ t ≤ 4, zero for 4 < t ≤ 10, and −2 m/s² for t > 10 until it comes to rest.
- (a) Sketch the velocity-time graph and find the time at which the particle comes to rest. [3]
- (b) Find the total distance travelled. [2]
Show answer
- (a) rises to 12 m/s at t = 4, horizontal to t = 10, falls to 0 at t = 16
v = 12 at t = 4 B1; deceleration time 6 s M1; t = 16 A1
- (b) 132 m
24 + 72 + 36 M1 A1
Car A travels along a straight road at a constant speed of 20 m/s. At t = 0 it passes car B, which is at rest. B immediately accelerates uniformly at 2.5 m/s² until it reaches 30 m/s, and then continues at 30 m/s.
- (a) Find the time at which B catches up with A. [4]
- (b) Find the distance from the starting point at which this happens. [1]
Show answer
- (a) t = 18 s
B reaches 30 m/s at t = 12 having travelled 180 m B1; A has 240 m, so not yet B1; 180 + 30(t − 12) = 20t M1; t = 18 A1
- (b) 360 m
B1
A car in Peshawar starts from rest and accelerates uniformly to a maximum speed, then immediately decelerates uniformly to rest. The whole journey takes 60 s and covers 900 m. The magnitude of the deceleration is half the magnitude of the acceleration.
- (a) Find the maximum speed of the car. [2]
- (b) Find the acceleration and the deceleration of the car. [4]
Show answer
- (a) 30 m/s
area = ½ × 60 × V = 900 M1; V = 30 A1
- (b) 1.5 m/s² and 0.75 m/s²
deceleration time = 2 × acceleration time M1; t₁ = 20, t₂ = 40 A1; 30/20 = 1.5 A1; 30/40 = 0.75 A1
A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity is v m/s, where v = t² − 6t + 8.
- (a) Find the minimum velocity of the particle and state the interval of time during which it moves towards O. [3]
- (b) Find the total distance travelled for 0 ≤ t ≤ 5. [4]
Show answer
- (a) minimum −1 m/s at t = 3; moves towards O (backwards) for 2 < t < 4
a = 2t − 6 = 0, t = 3 M1; v = −1 A1; 2 < t < 4 B1
- (b) 28/3 m
s = t³/3 − 3t² + 8t M1; s(2) = 20/3, s(4) = 16/3, s(5) = 20/3 A1 A1; 20/3 + 4/3 + 4/3 = 28/3 A1
