Free access until 31 December 2026 · Physics and Maths questions and notes, updated continuously · Build a test · Get an email before anything changes

ExaminerPrep › Question bank › IAL Maths › M1.2 Kinematics in a straight line

Edexcel International A Level Maths · M1 · Mechanics 1

Kinematics in a straight line

28 free, high-quality past-paper-style questions on Kinematics in a straight line for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 238 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-M1-2-B-01 M1 · ASEasy6 marks

A bus is travelling along a straight horizontal road at 16 m s−1 when the driver applies the brakes. The bus decelerates uniformly and comes to rest after travelling 64 m. The bus is modelled as a particle.

(a)

Find the deceleration of the bus.

(2)
(b)

Find the time taken for the bus to come to rest.

(2)
(c)

Find the distance travelled by the bus in the last 2 s before it stops.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
v2 = u2 + 2as: 0 = 162 + 2a × 64M1
a = −2, so the deceleration is 2 m s−2A1
Part (b)
v = u + at: 0 = 16 − 2t, or s = ½(u + v)t: 64 = 8tM1
t = 8 s (ft from (a))A1
Part (c)
speed 2 s before stopping: 16 − 2 × 6 = 4 m s−1, or use s = vt − ½at2 with v = 0, t = 2M1
s = ½(4 + 0) × 2 = 4 mA1

Exam tip: For the last few seconds of a motion that ends at rest, s = vt − ½at2 with v = 0 gives the distance at once, without finding the earlier speed.

2IALM-M1-2-B-02 M1 · ASEasy6 marks

A coin is dropped from rest from a footbridge and falls vertically into the river below. It reaches the water 2.5 s after it is released. The particle model is used, air resistance is ignored and g = 9.8 m s−2.

(a)

Find the height of the point of release above the water.

(2)
(b)

Find the speed of the coin as it reaches the water.

(2)
(c)

Find the time taken for the coin to fall the first half of this height.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
down positive: s = ut + ½at2 = 0 + ½ × 9.8 × 2.52M1
s = 30.625, so the height is 30.6 m (accept 31 m)A1
Part (b)
v = u + at = 0 + 9.8 × 2.5M1
= 24.5 m s−1A1
Part (c)
½ × 30.625 = 4.9t2M1
t = 1.7677…, so 1.77 s (accept 1.8 s)A1

Common error: Halving the time to find when half the height has been fallen. Distance grows with t2, so half the height is reached after 2.5/√2 s, not 1.25 s.

3IALM-M1-2-B-03 M1 · ASEasy6 marks

A jogger runs along a straight path that starts at her front gate H. Her displacement–time graph consists of three straight line segments joining the points (0, 0), (300, 900), (420, 900) and (720, 0), where t is the time in seconds and s is her displacement from H in metres.

300420720900t (s)s (m)
Figure: displacement–time graph of the jogger, drawn from the points given.
(a)

Find her velocity during the first 300 s.

(1)
(b)

Describe her motion for 300 ≤ t ≤ 420.

(1)
(c)

Find her velocity for 420 ≤ t ≤ 720.

(2)
(d)

Find her average speed for the whole 720 s, and state her average velocity for the whole 720 s, giving a reason.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = 900 ÷ 300 = 3 m s−1 (away from H)B1
Part (b)
she is at rest (stationary), 900 m from HB1
Part (c)
gradient = (0 − 900) ÷ (720 − 420)M1
= −3 m s−1, i.e. 3 m s−1 back towards HA1
Part (d)
average speed = (900 + 900) ÷ 720 = 2.5 m s−1B1
average velocity = 0, because her final displacement from H is zeroB1

Exam tip: On a displacement–time graph the gradient is the velocity. A negative gradient means motion back towards the origin — give the direction in words as well as the sign.

4IALM-M1-2-B-04 M1 · ASEasy5 marks

A speed skater moves along a straight section of an ice track with constant acceleration 0.8 m s−2. During a period of 6.0 s her speed increases to 11.4 m s−1. The skater is modelled as a particle.

(a)

Find the distance she travels during the 6.0 s.

(2)
(b)

Find her speed at the start of the 6.0 s period.

(2)
(c)

Find the time after the start of the period at which her speed is 9.0 m s−1.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
s = vt − ½at2 = 11.4 × 6.0 − ½ × 0.8 × 6.02M1
= 68.4 − 14.4 = 54 mA1
Part (b)
v = u + at: 11.4 = u + 0.8 × 6.0M1
u = 6.6 m s−1A1
Part (c)
(9.0 − 6.6) ÷ 0.8 = 3.0 s (ft from (b))B1

Exam tip: When the final speed is given but not the initial one, s = vt − ½at2 is the formula that leaves out u. Check your answer afterwards with s = ½(u + v)t.

5IALM-M1-2-FB-01 M1 · ASEasy7 marks

A lift moves vertically upwards from rest at the ground floor of a building and comes to rest at a higher floor. Figure 1 shows the velocity–time graph of the lift for the whole journey. The lift is modelled as a particle.

01234012345678910111213141516t / sv / m s⁻¹
(a)

Find the acceleration of the lift during the first 2 s.

(1)
(b)

Find the deceleration of the lift during the final stage of the journey.

(1)
(c)

Find the height through which the lift rises.

(3)
(d)

Find the time after the start at which the lift is 18 m above the ground floor.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = 3 ÷ 2 = 1.5 m s−2B1
Part (b)
3 ÷ (15 − 12) = 1 m s−2B1
Part (c)
height = area under the graph: ½ × 2 × 3 + 10 × 3 + ½ × 3 × 3, or trapezium ½(10 + 15) × 3M1
= 3 + 30 + 4.5A1
= 37.5 mA1
Part (d)
3 m are gained in the first 2 s, so 18 m is reached at constant speed: 3 + 3(t − 2) = 18M1
t = 7 sA1

Exam tip: The area under a velocity–time graph is the displacement. A trapezium-shaped graph can be done in one step: ½ × (sum of the parallel sides) × height, with parallel sides 10 s and 15 s here.

6IALM-M1-2-FB-02 M1 · ASEasy6 marks

A robot cleaner moves along a straight corridor. Figure 1 shows the graph of its displacement s metres from a fixed point O in the corridor against time t seconds, for 0 ≤ t ≤ 20. The graph consists of four straight line segments. The robot is modelled as a particle.

−6−4−20246810121402468101214161820t / ss / mO
(a)

Find the velocity of the robot for 0 < t < 5.

(1)
(b)

Find the velocity of the robot for 8 < t < 16.

(2)
(c)

Find the time, for 0 < t < 20, at which the robot passes through O.

(1)
(d)

Find the total distance travelled by the robot for 0 ≤ t ≤ 20, and its average speed over this time.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (12 − 2) ÷ 5 = 2 m s−1B1
Part (b)
gradient = (−4 − 12) ÷ (16 − 8)M1
= −2 m s−1, i.e. 2 m s−1 in the negative directionA1
Part (c)
12 − 2(t − 8) = 0 ⇒ t = 14 (allow read from the graph)B1
Part (d)
distance = 10 + 0 + 16 + 4 = 30 mB1
average speed = 30 ÷ 20 = 1.5 m s−1 (ft)B1

Common error: Using the change in displacement, 0 − 2 = −2 m, as the distance travelled. Distance adds the length of every leg, whichever way the robot is moving.

7IALM-M1-2-FB-03 M1 · ASEasy7 marks

Figure 1 shows a ball thrown vertically downwards with speed 3.5 m s−1 from a window W. The ball falls past the level of a balcony B, without touching it, and hits the horizontal ground. The heights of W and B above the ground are shown in the figure. The particle model is used, air resistance is ignored and g = 9.8 m s−2.

groundWB3.5 m s⁻¹26.6 m12.6 m
(a)

Show that the ball hits the ground 2 s after it is thrown.

(3)
(b)

Find the speed of the ball as it hits the ground.

(2)
(c)

Find the time taken for the ball to fall from W to the level of B.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
down positive: 26.6 = 3.5t + 4.9t2M1
7t2 + 5t − 38 = 0 ⇒ (7t + 19)(t − 2) = 0, or the formulaM1
t = 2, the negative root rejected (given answer, fully shown)A1
Part (b)
v = 3.5 + 9.8 × 2M1
= 23.1 m s−1A1
Part (c)
distance fallen = 26.6 − 12.6 = 14 m: 14 = 3.5t + 4.9t2M1
t = 1.3705…, so 1.37 sA1

Common error: Using 12.6 m, the height of B, as the distance fallen. Displacement is measured from the point of release, so the ball falls 26.6 − 12.6 = 14 m to reach B.

8IALM-M1-2-FB-04 M1 · ASEasy6 marks

A tram moves along the straight track shown in Figure 1. It passes point A and accelerates uniformly until it passes point B. It then decelerates uniformly from B and comes to rest at point C. Its speeds at A and B and the distances AB and BC are shown in the figure. The tram is modelled as a particle.

ABC8 m s⁻¹16 m s⁻¹at rest120 m80 mdirection of motion →
(a)

Find the acceleration of the tram between A and B.

(2)
(b)

Find the total time taken by the tram to travel from A to C.

(3)
(c)

Find the average speed of the tram from A to C.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
162 = 82 + 2a × 120M1
a = 0.8 m s−2A1
Part (b)
A to B: 120 = ½(8 + 16)t ⇒ t = 10 sM1
B to C: 80 = ½(16 + 0)t ⇒ t = 10 sM1
total = 20 sA1
Part (c)
200 ÷ 20 = 10 m s−1 (ft)B1

Exam tip: When both end speeds and the distance are known, s = ½(u + v)t gives the time at once, without finding the acceleration first.

9IALM-M1-2-B-05 M1 · ASMedium8 marks

A pebble is fired vertically upwards from a catapult at ground level with speed 17.5 m s−1. The particle model is used, air resistance is ignored and g = 9.8 m s−2.

(a)

Find the greatest height above the ground reached by the pebble.

(2)
(b)

Find the speed of the pebble when it is 10 m above the ground.

(2)
(c)

Find the length of time for which the pebble is at least 10 m above the ground.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
up positive, at the top v = 0: 0 = 17.52 − 2 × 9.8 × sM1
s = 15.625, so 15.6 m (accept 16 m)A1
Part (b)
v2 = 17.52 − 2 × 9.8 × 10 = 110.25M1
v = 10.5 m s−1A1
Part (c)
10 = 17.5t − 4.9t2M1
4.9t2 − 17.5t + 10 = 0, solved by formula or calculatorM1
t = 0.714… and 2.857…A1
time = 2.857 − 0.714 = 2.14 s (exactly 15/7 s)A1
Alternative: by symmetry the pebble passes 10 m going up and coming down at 10.5 m s−1; −10.5 = 10.5 − 9.8t gives the same 2.14 s (marks as above)—

Exam tip: A quadratic for the time at a given height has two positive roots: the way up and the way down. Their difference is the time spent above that height.

10IALM-M1-2-B-06 M1 · ASMedium8 marks

A camera drone rises vertically from rest on the ground. It accelerates uniformly for 4 s until its speed is 6 m s−1. It then rises at a constant speed of 6 m s−1 for T seconds, and then decelerates uniformly, coming to rest after a further 3 s. It is then 90 m above the ground. The drone is modelled as a particle.

(a)

Sketch a velocity–time graph for the motion of the drone.

(2)
(b)

Find the value of T.

(3)
(c)

Find the deceleration of the drone in the final stage.

(1)
(d)

Find the time after leaving the ground at which the drone is 50 m above the ground.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
trapezium shape: straight line from the origin, horizontal segment, straight line back to v = 0B1
6, 4, T and 3 (or 4 + T and 7 + T) marked on the axesB1
Part (b)
area under the graph = 90: ½ × 4 × 6 + 6T + ½ × 3 × 6 = 90, or ½(T + T + 7) × 6 = 90M1
12 + 6T + 9 = 90A1
T = 11.5A1
Part (c)
6 ÷ 3 = 2 m s−2B1
Part (d)
12 m are gained in the first 4 s, so 50 m is reached at constant speed: 12 + 6(t − 4) = 50M1
t = 31/3 = 10.3 sA1

Model sketch

415.518.56t (s)v (m s⁻¹)
Figure: velocity–time graph for the drone (T = 11.5); not to scale.

Exam tip: Put the unknown T on your sketch before writing the area equation — the sketch then shows you which pieces of area to add.

11IALM-M1-2-B-08 M1 · ASMedium7 marks

Two cyclists, A and B, ride towards each other along a straight cycle path. At time t = 0 they are 200 m apart. A starts from rest and accelerates uniformly at 0.4 m s−2. B rides at a constant speed of 4 m s−1 throughout. The cyclists are modelled as particles and A is still accelerating when they meet.

AB0.4 m s⁻²from rest4 m s⁻¹200 m
Figure: the cyclists at t = 0; not to scale.
(a)

Show that they meet when t2 + 20t − 1000 = 0.

(3)
(b)

Find the time at which they meet.

(2)
(c)

Find the distance from B's starting point to the point where they meet, and the speed of A at that moment.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
distance travelled by A: ½ × 0.4t2 = 0.2t2B1
distance travelled by B: 4tB1
0.2t2 + 4t = 200 ⇒ t2 + 20t − 1000 = 0 (given answer, fully shown)A1
Part (b)
t = −10 + √1100 (positive root)M1
t = 23.2 sA1
Part (c)
4 × 23.17 = 92.7 mB1
0.4 × 23.17 = 9.27 m s−1 (accept 9.3 m s−1)B1

Common error: Setting the two displacements equal when the cyclists move towards each other. Here the distances they travel must add up to the 200 m gap.

12IALM-M1-2-B-09 M1 · ASMedium9 marks

An automated cart moves along a straight track in a warehouse. At time t = 0 its velocity is 0.5 m s−1. Its acceleration–time graph shows that its acceleration is 0.25 m s−2 for 0 < t < 6, zero for 6 < t < 26, and −0.4 m s−2 from t = 26 until the cart comes to rest at t = T. The cart is modelled as a particle.

626T−0.40.25t (s)a (m s⁻²)
Figure: acceleration–time graph of the cart; not to scale.
(a)

Find the velocity of the cart at t = 6.

(2)
(b)

Find the value of T.

(2)
(c)

Sketch the velocity–time graph for 0 ≤ t ≤ T.

(2)
(d)

Find the total distance travelled by the cart.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
change in velocity = area under the acceleration–time graph = 0.25 × 6 = 1.5M1
v = 0.5 + 1.5 = 2 m s−1A1
Part (b)
time to stop = 2 ÷ 0.4 = 5 sM1
T = 26 + 5 = 31A1
Part (c)
three straight segments: rising, horizontal, falling to zeroB1
starting at 0.5, with 2, 6, 26 and 31 markedB1
Part (d)
sum of areas: ½(0.5 + 2) × 6 + 2 × 20 + ½ × 2 × 5M1
= 7.5 + 40 + 5A1
= 52.5 mA1

Model sketch

626310.52t (s)v (m s⁻¹)
Figure: velocity–time graph of the cart (T = 31); not to scale.

Exam tip: The area under an acceleration–time graph is the change in velocity, not the velocity itself. Add the initial velocity.

13IALM-M1-2-B-10 M1 · ASMedium8 marks

A hot-air balloon is rising vertically with constant speed 4.9 m s−1. When the basket is 58.8 m above the ground, a sandbag is released from the basket. The sandbag then moves freely under gravity. The particle model is used, air resistance is ignored and g = 9.8 m s−2.

4.9 m s⁻¹58.8 m
Figure: the balloon rising at the moment the sandbag is released; not to scale.
(a)

Find the time taken for the sandbag to reach the ground.

(3)
(b)

Find the speed of the sandbag as it reaches the ground.

(2)
(c)

Find the greatest height of the sandbag above the ground.

(2)
(d)

Find the height of the basket above the ground at the instant the sandbag reaches the ground.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
up positive: u = +4.9, a = −9.8, s = −58.8B1
−58.8 = 4.9t − 4.9t2M1
t2 − t − 12 = 0 ⇒ t = 4 sA1
Part (b)
v = 4.9 − 9.8 × 4 = −34.3M1
speed = 34.3 m s−1A1
Part (c)
further rise: 0 = 4.92 − 2 × 9.8 × s ⇒ s = 1.225 mM1
greatest height = 58.8 + 1.225 = 60.0 m (accept 60 m)A1
Part (d)
58.8 + 4.9 × 4 = 78.4 mB1

Common error: Taking the initial velocity of the sandbag as zero. At release it has the balloon's velocity, 4.9 m s−1 upwards, so it rises a little before it falls.

14IALM-M1-2-B-11 M1 · ASMedium8 marks

A delivery van is travelling along a straight road at 18 m s−1 when the driver sees a fallen branch lying across the road 45 m ahead. The van continues at 18 m s−1 for a reaction time of 0.7 s, and then the driver brakes. While braking, the van decelerates uniformly at 4.5 m s−2. The van is modelled as a particle.

(a)

Show that the van cannot stop before it reaches the branch.

(3)
(b)

Find the speed of the van when it reaches the branch.

(3)
(c)

Find the greatest reaction time that would allow the van to stop just before the branch, with the same deceleration.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
distance during reaction time = 18 × 0.7 = 12.6 mB1
braking distance: 0 = 182 − 2 × 4.5 × s ⇒ s = 36 mM1
12.6 + 36 = 48.6 m > 45 m, so it cannot stop in timeA1
Part (b)
braking distance available = 45 − 12.6 = 32.4 mM1
v2 = 182 − 2 × 4.5 × 32.4 = 32.4M1
v = 5.69 m s−1 (accept 5.7)A1
Part (c)
18tr + 36 = 45M1
tr = 0.5 sA1

Exam tip: During the reaction time there is no braking: the vehicle moves at constant speed. Treat it as a separate first stage with a = 0.

15IALM-M1-2-FB-05 M1 · ASMedium7 marks

An electric go-kart moves along a straight track. It starts from rest at time t = 0. Figure 1 shows its acceleration–time graph for 0 ≤ t ≤ 14, where t is in seconds. The go-kart is modelled as a particle.

−1.5−1−0.500.511.522.530123456789101112131415t / sa / m s⁻²
(a)

Find the speed of the go-kart at t = 4 and at t = 10.

(2)
(b)

Find the total distance travelled by the go-kart for 0 ≤ t ≤ 14.

(3)
(c)

Find the two times at which the speed of the go-kart is 7 m s−1.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
change in velocity = area under the graph: 2.5 × 4 = 10, so 10 m s−1 at t = 4B1
10 − 1 × 6 = 4 m s−1 at t = 10B1
Part (b)
speed at t = 14: 4 − 0.5 × 4 = 2 m s−1B1
distance = ½ × 4 × 10 + ½(10 + 4) × 6 + ½(4 + 2) × 4M1
= 20 + 42 + 12 = 74 mA1
Part (c)
2.5t = 7 ⇒ t = 2.8B1
10 − (t − 4) = 7 ⇒ t = 7B1

Common error: Reading a negative acceleration as motion backwards. Between t = 4 and t = 14 the go-kart is still moving forwards — it is slowing down, and its velocity stays positive.

16IALM-M1-2-FB-06 M1 · ASMedium8 marks

A puck is projected up a straight ramp from a point A at the bottom of the ramp. It moves along a line of greatest slope with constant acceleration. Figure 1 shows its velocity–time graph for the first 6 s, taking the direction up the ramp as positive. The ramp is long enough for the puck not to reach the top. The puck is modelled as a particle.

−4−3−2−10123456701234567t / sv / m s⁻¹end of motion shown
(a)

Find the acceleration of the puck.

(1)
(b)

Find the greatest distance of the puck from A.

(2)
(c)

Find the displacement of the puck from A at t = 6.

(2)
(d)

Find the time at which the puck returns to A, and its speed at that instant.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (−3 − 6) ÷ 6 = −1.5, i.e. 1.5 m s−2 down the rampB1
Part (b)
v = 0 at t = 4 (from the graph or 0 = 6 − 1.5t)M1
area = ½ × 4 × 6 = 12 mA1
Part (c)
12 − ½ × 2 × 3, or s = vt − ½at2 = (−3)(6) + 0.75 × 36M1
= 9 m up the ramp from AA1
Part (d)
0 = 6t − 0.75t2M1
t = 8 s (t = 0 rejected)A1
v = 6 − 1.5 × 8 = −6, so speed 6 m s−1A1

Exam tip: Area below the time axis is displacement in the negative direction. Subtract it to find the displacement; add it to find the distance travelled.

17IALM-M1-2-FB-07 M1 · ASMedium8 marks

A ball is thrown vertically upwards from a point on the top of a wall and later falls past the wall to the ground. Figure 1 shows the graph of its height h metres above the ground against the time t seconds after it is thrown, until it reaches the ground. The highest point (1, 19.6) and the point (3, 0) are marked on the graph. The particle model is used, air resistance is ignored and g = 9.8 m s−2.

024681012141618202200.511.522.533.5t / sh / m(1, 19.6)(3, 0)
(a)

Use the graph to find the speed with which the ball is thrown.

(2)
(b)

Show that the height of the wall is 14.7 m.

(2)
(c)

Find the speed of the ball as it reaches the ground.

(2)
(d)

Find the length of time for which the ball is more than 17.15 m above the ground.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
at the highest point v = 0 when t = 1: 0 = u − 9.8 × 1M1
u = 9.8 m s−1A1
Part (b)
rise from the wall to the top: s = vt − ½at2 = 0 + 4.9 × 12 = 4.9 m, or the 3 s motion: s = 9.8 × 3 − 4.9 × 32 = −14.7M1
19.6 − 4.9 = 14.7 m (given answer)A1
Part (c)
v = 9.8 − 9.8 × 3 = −19.6, or a 19.6 m fall from rest in 2 sM1
speed = 19.6 m s−1A1
Part (d)
14.7 + 9.8t − 4.9t2 = 17.15 ⇒ t2 − 2t + 0.5 = 0, or 2.45 = 4.9t2 measured from the topM1
t = 1 ± √0.5, so the time is 2√0.5 = √2 = 1.41 sA1

Exam tip: The motion is symmetrical about the highest point: the ball takes as long to rise from a given height to the top as to fall back to that height. Half the time above 17.15 m is the time to fall 2.45 m from rest.

18IALM-M1-2-FB-08 M1 · ASMedium8 marks

A ball is released from rest at time t = 0 and falls vertically onto a horizontal floor. It bounces and rises vertically, and then falls back to the floor. Figure 1 shows its velocity–time graph for 0 ≤ t ≤ 2.4, taking downwards as positive; the marked points have the coordinates shown. The time for which the ball is in contact with the floor is negligible. The particle model is used, air resistance is ignored and g = 9.8 m s−2.

−8−6−4−2024681000.511.522.5t / sv / m s⁻¹(1, 9.8)(1, −6.86)(1.7, 0)(2.4, 6.86)downwards positive
(a)

State the value of the gradient of each sloping line, and what it represents.

(1)
(b)

Find the height above the floor from which the ball is released.

(2)
(c)

Find the greatest height of the ball above the floor after the bounce.

(2)
(d)

Find the total distance travelled by the ball for 0 ≤ t ≤ 2.4.

(2)
(e)

Find the speed of the ball 0.5 s after it leaves the floor, and state its direction of motion.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
9.8 m s−2: the acceleration due to gravity, downwards, the same before and after the bounceB1
Part (b)
area of the first triangle: ½ × 1 × 9.8M1
= 4.9 mA1
Part (c)
area between t = 1 and t = 1.7: ½ × 0.7 × 6.86M1
= 2.401, so 2.40 mA1
Part (d)
4.9 + 2 × 2.401M1
= 9.702, so 9.70 mA1
Part (e)
v = −6.86 + 9.8 × 0.5 = −1.96: 1.96 m s−1 upwardsB1

Exam tip: The vertical jump at the bounce is where the velocity changes sign. Negative velocity here means moving upwards, so the area of the part below the axis is the height of the rebound.

19IALM-M1-2-FB-09 M1 · ASMedium9 marks

In a fitness test a runner starts from rest on a line A, runs in a straight line to a parallel line B, and then runs straight back to A, where she stops. She is instantaneously at rest at B when t = 6. Figure 1 shows her speed–time graph for the whole test, 0 ≤ t ≤ 13, where t is in seconds. The runner is modelled as a particle.

012345601234567891011121314t / sspeed / m s⁻¹
(a)

Show that the distance between the lines is 20 m.

(2)
(b)

Find her deceleration as she approaches B.

(1)
(c)

Find her distance from A at t = 10.

(2)
(d)

Taking the direction from A to B as positive, sketch the velocity–time graph for 0 ≤ t ≤ 13.

(2)
(e)

Find her average speed and her average velocity for the whole test.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
area for 0 ≤ t ≤ 6: ½ × 2 × 5 + 2 × 5 + ½ × 2 × 5, or ½(2 + 6) × 5M1
= 20 m (given answer)A1
Part (b)
5 ÷ 2 = 2.5 m s−2B1
Part (c)
distance run back since t = 6: ½ × 2 × 4 + 2 × 4 = 12 mM1
20 − 12 = 8 mA1
Part (d)
0 ≤ t ≤ 6 the same as the speed–time graph, above the axisB1
6 ≤ t ≤ 13 reflected below the axis: down to −4 at t = 8, level to t = 11, back to 0 at t = 13B1
Part (e)
average speed = 40 ÷ 13 = 3.08 m s−1B1
average velocity = 0, as she finishes where she startedB1

Common error: Treating the speed–time graph as a velocity–time graph and adding the two areas as displacement. Speed is never negative, so the second leg must be counted as motion back towards A.

20IALM-M1-2-FB-10 M1 · ASMedium8 marks

Figure 1 shows a diver on a springboard. She leaves the end of the board moving vertically upwards with speed 4.9 m s−1 and moves freely under gravity until she enters the water. In the water she moves vertically downwards with constant deceleration and comes to rest at the depth shown. The height of the end of the board above the water surface is shown in the figure. The diver is modelled as a particle moving in a vertical line that does not meet the board; in the air, air resistance is ignored and g = 9.8 m s−2.

water surface4.9 m s⁻¹comes to rest3.0 m2.6 m
(a)

Find the greatest height of the diver above the water surface.

(2)
(b)

Find the speed of the diver as she enters the water.

(2)
(c)

Find the time from leaving the board to entering the water.

(2)
(d)

Find the deceleration of the diver in the water.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
up positive: 0 = 4.92 − 2 × 9.8 × s ⇒ s = 1.225 mM1
3.0 + 1.225 = 4.225 m (accept 4.2 m or 4.23 m)A1
Part (b)
v2 = 4.92 + 2 × 9.8 × 3.0 = 82.81, or a 4.225 m fall from restM1
v = 9.1 m s−1A1
Part (c)
−9.1 = 4.9 − 9.8t, or −3.0 = 4.9t − 4.9t2M1
t = 14/9.8 = 10/7, so 1.43 sA1
Part (d)
0 = 9.12 − 2a × 2.6M1
a = 15.925, so 15.9 m s−2A1

Exam tip: Take upwards as positive for the flight: the water surface is then at displacement −3.0 m from the board. Start a new stage at the surface, using the entry speed as the new initial speed.

21IALM-M1-2-B-07 M1 · ASHard12 marks

A particle P moves along a straight line with constant acceleration. At time t = 0, P passes through a fixed point O on the line, moving with speed 9 m s−1 in the positive direction. The acceleration of P is 2 m s−2 in the negative direction. The displacement of P from O at time t seconds is s metres.

(a)

Find the time at which P is instantaneously at rest.

(1)
(b)

Find the displacement of P from O when t = 7.

(2)
(c)

Find the total distance travelled by P in the interval 0 ≤ t ≤ 7.

(3)
(d)

Sketch the speed–time graph of P for 0 ≤ t ≤ 7.

(2)
(e)

Find all the values of t for which P is at a distance of 8 m from O.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
0 = 9 − 2t ⇒ t = 4.5B1
Part (b)
s = 9 × 7 − ½ × 2 × 72M1
= 14 mA1
Part (c)
displacement at t = 4.5: 9 × 4.5 − 4.52 = 20.25 mM1
distance back from t = 4.5 to 7: 20.25 − 14 = 6.25 mM1
total = 26.5 mA1
Part (d)
straight line from (0, 9) down to (4.5, 0)B1
straight line from (4.5, 0) up to (7, 5), the graph never below the axisB1
Part (e)
s = 8: 9t − t2 = 8 ⇒ (t − 1)(t − 8) = 0M1
t = 1 and 8A1
s = −8: t2 − 9t − 8 = 0M1
t = (9 + √113)/2 = 9.82 (3 s.f.), the negative root rejectedA1

Model sketch

4.5759t (s)speed (m s⁻¹)
Figure: speed–time graph of P for 0 ≤ t ≤ 7; not to scale.

Exam tip: “At a distance of 8 m from O” includes points on both sides of O: solve for s = 8 and for s = −8, then reject any negative time.

22IALM-M1-2-B-12 M1 · ASHard12 marks

A motorcycle is at rest at a junction on a straight road. At time t = 0 a bus passes the junction, travelling along the road at a constant speed of 16 m s−1. At t = 3 s the motorcycle sets off after the bus, moving with constant acceleration 2 m s−2 until it reaches its top speed of 28 m s−1. It then continues at 28 m s−1. Both vehicles are modelled as particles moving along the same straight line.

(a)

Find the value of t when the motorcycle reaches its top speed.

(1)
(b)

Show that the motorcycle has not caught up with the bus by this time.

(3)
(c)

Find the time at which the motorcycle overtakes the bus, and its distance from the junction at that moment.

(4)
(d)

Sketch, on the same axes, velocity–time graphs for the bus and the motorcycle for 0 ≤ t ≤ 25.

(2)
(e)

Find the greatest distance between the bus and the motorcycle before the motorcycle overtakes.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
3 + 28 ÷ 2 = 17B1
Part (b)
motorcycle: ½ × 2 × 142 = 196 mM1
bus: 16 × 17 = 272 mM1
196 < 272, so the bus is still aheadA1
Part (c)
for t ≥ 17: 196 + 28(t − 17) = 16tM1
12t = 280A1
t = 70/3 = 23.3 sA1
distance = 16 × 70/3 = 373 mB1
Part (d)
bus: horizontal line at 16B1
motorcycle: zero until t = 3, straight line up to 28 at t = 17, then horizontalB1
Part (e)
the gap is greatest when the speeds are equal: 2(t − 3) = 16 ⇒ t = 11M1
gap = 16 × 11 − ½ × 2 × 82 = 176 − 64 = 112 mA1

Model sketch

317251628t (s)v (m s⁻¹)busmotorcycle
Figure: velocity–time graphs of the bus and the motorcycle; not to scale.

Common error: Solving (t − 3)2 = 16t for the overtaking time. Its root, about 21.6 s, assumes the motorcycle is still accelerating after t = 17, which it is not.

23IALM-M1-2-B-13 M1 · ASHard11 marks

Ball A is projected vertically upwards from a point X on the ground with speed 24.5 m s−1. At the same instant, ball B is released from rest at a point H metres vertically above X. The balls move in the same vertical line. Both balls are modelled as particles, air resistance is ignored and g = 9.8 m s−2. In parts (a) to (c), H = 49.

ABX24.5 m s⁻¹(released from rest)H m
Figure: ball A projected upwards from X and ball B released at height H m above X; not to scale.
(a)

Show that the balls collide 2 s after they are set in motion.

(3)
(b)

Find the height above the ground at which they collide.

(2)
(c)

Find the velocity of each ball just before they collide.

(3)
(d)

For a general value of H, find the set of values of H for which the balls collide while A is still moving upwards.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
height of A above X: 24.5t − 4.9t2B1
height of B above X: 49 − 4.9t2B1
equal heights: 24.5t = 49 ⇒ t = 2 (given answer)A1
Part (b)
24.5 × 2 − 4.9 × 22M1
= 29.4 mA1
Part (c)
A: v = 24.5 − 9.8 × 2M1
= 4.9 m s−1 upwardsA1
B: 9.8 × 2 = 19.6 m s−1 downwardsB1
Part (d)
the t2 terms cancel again: collision at t = H/24.5M1
A is moving upwards while t < 24.5/9.8 = 2.5M1
H/24.5 < 2.5 ⇒ 0 < H < 61.25A1

Exam tip: Measure both heights from the same origin (here X) with the same positive direction. The −4.9t2 terms then cancel, because both balls have the same acceleration.

24IALM-M1-2-B-14 M1 · ASHard12 marks

A test vehicle on a straight track starts from rest and accelerates uniformly at a m s−2 until its speed is 30 m s−1. It then travels at 30 m s−1 for a time, and then decelerates uniformly at 2a m s−2 until it comes to rest. The whole journey takes 100 s and covers a distance of 2.55 km. The vehicle is modelled as a particle.

(a)

Sketch a velocity–time graph for the journey.

(2)
(b)

Show that the time for which the vehicle travels at 30 m s−1 is (100 − 45/a) seconds.

(2)
(c)

Find the value of a.

(4)
(d)

Find the two times at which the speed of the vehicle is 12 m s−1.

(2)
(e)

Find the time taken for the vehicle to travel the first 1 km.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
trapezium starting and ending on the time axis, with the deceleration side steeperB1
30 and 100 markedB1
Part (b)
accelerating for 30/a s and decelerating for 30/(2a) = 15/a sB1
100 − 30/a − 15/a = 100 − 45/a (given answer)A1
Part (c)
area = 2550: ½(100 + 100 − 45/a) × 30 = 2550, or the three separate areasM1
15(200 − 45/a) = 2550A1
200 − 45/a = 170, so 45/a = 30M1
a = 1.5A1
Part (d)
12 ÷ 1.5 = 8 sB1
deceleration starts at 100 − 10 = 90 s: 90 + (30 − 12) ÷ 3 = 96 sB1
Part (e)
first 20 s: ½ × 20 × 30 = 300 m, so 300 + 30(t − 20) = 1000M1
t = 130/3 = 43.3 sA1

Model sketch

209010030t (s)v (m s⁻¹)
Figure: velocity–time graph for the journey (a = 1.5); not to scale.

Exam tip: With an unknown acceleration, write each time as change in speed ÷ acceleration in terms of a, then the area of the trapezium gives one equation in a.

25IALM-M1-2-FB-11 M1 · ASHard11 marks

A coach passes a junction J at time t = 0, travelling along a straight road and slowing down for roadworks ahead. At the same instant an ambulance starts from rest at J and moves along the same road in the same direction. Figure 1 shows the velocity–time graphs of the coach and the ambulance for 0 ≤ t ≤ 25, where t is in seconds. Both vehicles are modelled as particles.

02468101214161820220510152025t / sv / m s⁻¹coachambulance
(a)

Find the time at which the two vehicles have the same speed, and this speed.

(2)
(b)

Find the greatest distance between the vehicles before the ambulance overtakes the coach.

(3)
(c)

Show that the ambulance has not overtaken the coach by t = 10.

(2)
(d)

Find the time at which the ambulance overtakes the coach, and the distance from J at which this happens.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
coach: v = 15 − 0.5t; ambulance: v = 2tM1
t = 6, speed 12 m s−1 (allow read from the graph)A1
Part (b)
the gap grows while the coach is faster, so it is greatest at t = 6M1
coach: 15 × 6 − 0.25 × 62 = 81 m; ambulance: ½ × 6 × 12 = 36 mM1
gap = 45 mA1
Part (c)
coach at t = 10: ½(15 + 10) × 10 = 125 mM1
ambulance: ½ × 10 × 20 = 100 m < 125 m, so not yetA1
Part (d)
for 10 ≤ t ≤ 20: 100 + 20(t − 10) = 15t − 0.25t2M1
t2 + 20t − 400 = 0A1
t = −10 + √500 = 12.36…, so 12.4 sA1
distance = 20 × 12.36 − 100 = 147 mA1

Common error: Using ½ × 2t2 for the ambulance after t = 10. It reaches 20 m s−1 at t = 10 and then moves at constant speed, so its distance becomes 100 + 20(t − 10).

26IALM-M1-2-FB-12 M1 · ASHard11 marks

Two boats, P and Q, move in the same direction along a straight canal. At time t = 0 they are in the positions shown in Figure 1, with the velocities shown. The engine of P has been switched off, so P decelerates uniformly at the rate shown until it comes to rest, and then remains at rest. Q moves at constant speed throughout. The boats are modelled as particles moving along parallel straight lines, and t is in seconds.

PQAB15 m s⁻¹1 m s⁻²3 m s⁻¹36 m
(a)

Show that P overtakes Q when t2 − 24t + 72 = 0, and find this value of t.

(4)
(b)

Find the greatest distance by which P is ahead of Q.

(3)
(c)

Find the time at which P comes to rest, and its distance from A at that time.

(2)
(d)

Find the time at which Q passes P.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
P: s = 15t − 0.5t2 from AB1
Q: s = 36 + 3t from AB1
15t − 0.5t2 = 36 + 3t ⇒ t2 − 24t + 72 = 0 (given answer)A1
t = 12 − 6√2 = 3.51 (the first time they are level)A1
Part (b)
greatest lead when their speeds are equal: 15 − t = 3 ⇒ t = 12M1
P: 180 − 72 = 108 m; Q: 36 + 36 = 72 mM1
lead = 36 mA1
Part (c)
0 = 15 − t ⇒ t = 15B1
152 ÷ 2 = 112.5 mB1
Part (d)
P is at rest 112.5 m from A: 36 + 3t = 112.5M1
t = 25.5A1

Common error: Taking the other root, 12 + 6√2 ≈ 20.5, as the time Q passes P. That root assumes P keeps decelerating and moves backwards after t = 15, which it does not.

27IALM-M1-2-FB-13 M1 · ASHard12 marks

A train travels along a straight track from rest at station P to rest at station Q, a distance of 7.5 km. Figure 1 is a sketch of its speed–time graph, showing its speeds in m s−1 and the duration in seconds of each of the five stages. The train travels at its greatest speed for T seconds, then slows to pass through a section of track with a speed limit, and each change of speed is at a constant rate. The train is modelled as a particle.

t / sspeed / m s⁻¹2412040T309040Not to scale
(a)

Find the acceleration of the train in the first stage and its deceleration in the last stage.

(2)
(b)

Find the length of the section of track with the speed limit.

(1)
(c)

Find the value of T.

(3)
(d)

Find the average speed of the train for the journey from P to Q.

(2)
(e)

On another day there is no speed limit. The train accelerates as before, travels at 24 m s−1 and then decelerates at 0.3 m s−2 directly to rest at Q. Find the time saved.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
24 ÷ 40 = 0.6 m s−2B1
12 ÷ 40 = 0.3 m s−2B1
Part (b)
12 × 90 = 1080 mB1
Part (c)
total area = 7500: ½ × 40 × 24 + 24T + ½(24 + 12) × 30 + 1080 + ½ × 40 × 12 = 7500M1
480 + 24T + 540 + 1080 + 240 = 7500A1
T = 215A1
Part (d)
total time = 40 + 215 + 30 + 90 + 40 = 415 sM1
7500 ÷ 415 = 18.07…, so 18.1 m s−1A1
Part (e)
deceleration takes 24 ÷ 0.3 = 80 s and covers ½ × 80 × 24 = 960 mM1
at 24 m s−1: (7500 − 480 − 960) ÷ 24 = 252.5 sM1
total time = 40 + 252.5 + 80 = 372.5 sA1
time saved = 415 − 372.5 = 42.5 sA1

Exam tip: Split the area under a multi-stage graph into one triangle, rectangle or trapezium per stage and write each area next to its stage before forming the equation for T.

28IALM-M1-2-FB-14 M1 · ASHard12 marks

Figure 1 shows the displacement–time graph for a particle R, which moves on a straight line with constant acceleration and is at the fixed point O on the line at time t = 0. Here s metres is the displacement of R from O at time t seconds, for 0 ≤ t ≤ 11. The points (2, 16) and (6, 24) lie on the curve.

−30−20−100102030024681012t / ss / m(2, 16)(6, 24)O
(a)

Find the initial velocity u m s−1 and the acceleration a m s−2 of R.

(4)
(b)

Find the greatest displacement of R from O in the positive direction, and the time at which it occurs.

(2)
(c)

Find the total distance travelled by R for 0 ≤ t ≤ 11.

(3)
(d)

Find the length of time, during 0 ≤ t ≤ 11, for which the speed of R is less than 4 m s−1.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
s = ut + ½at2 at t = 2: 16 = 2u + 2aM1
at t = 6: 24 = 6u + 18aM1
solving: a = −2A1
u = 10A1
Part (b)
v = 10 − 2t = 0 ⇒ t = 5M1
s = 50 − 25 = 25 mA1
Part (c)
s(11) = 110 − 121 = −11B1
distance = 25 + (25 + 11)M1
= 61 mA1
Part (d)
v = 10 − 2tB1
−4 < 10 − 2t < 4 ⇒ 3 < t < 7M1
length of time = 4 sA1

Common error: Giving the distance travelled as 11 m, the size of the final displacement. R turns round at t = 5, so the outward and return distances must be added.