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Edexcel International A Level Maths · M2 · Mechanics 2

Variable acceleration and vectors with calculus

28 free, high-quality past-paper-style questions on Variable acceleration and vectors with calculus 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 · 261 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-M2-2-B-01 M2 · A2Easy6 marks

A self-driving test vehicle moves along a straight test track. At time t seconds, its displacement from a fixed point O on the track is x metres, where
x = 0.5t3 − 6t2 + 18t,   0 ≤ t ≤ 7.
The vehicle is modelled as a particle.

(a)

Find an expression for the velocity of the vehicle at time t seconds.

(2)
(b)

Find the values of t at which the vehicle is instantaneously at rest.

(2)
(c)

Find the displacement of the vehicle from O when it first comes to rest.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
differentiates, at least one power of t reduced by oneM1
v = 1.5t2 − 12t + 18A1
Part (b)
sets v = 0 and solves: 1.5(t − 2)(t − 6) = 0M1
t = 2 and t = 6A1
Part (c)
substitutes the smaller value, t = 2, into xM1
x = 4 − 24 + 36 = 16 mA1

Common error: Substituting the time into v instead of x. The question asks for a displacement, so go back to the original expression for x.

2IALM-M2-2-B-02 M2 · A2Easy7 marks

A drone takes off from rest from level ground and rises vertically. At time t seconds after take-off its velocity is v m s−1 upwards, where
v = 2.4t − 0.3t2,   0 ≤ t ≤ 8.
The drone is modelled as a particle.

(a)

Find the acceleration of the drone when t = 2.

(2)
(b)

Find the greatest speed of the drone.

(2)
(c)

Find the height of the drone above the ground when t = 8.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
a = dv/dt = 2.4 − 0.6tM1
t = 2: a = 1.2 m s−2A1
Part (b)
a = 0 gives t = 4M1
v = 9.6 − 4.8 = 4.8 m s−1A1
Part (c)
integrates v, at least one power of t increased by oneM1
h = 1.2t2 − 0.1t3, constant zero since h = 0 when t = 0A1
h = 76.8 − 51.2 = 25.6 mA1

Exam tip: For “greatest speed”, set the acceleration to zero, then check that this value of t lies inside the interval and compare with the ends (here v = 0 at both ends).

3IALM-M2-2-B-03 M2 · A2Easy6 marks

A model boat moves on a pond. At time t seconds its position vector relative to a fixed point O is r metres, where
r = (3t2 − 2t)i + (5 + 4t − t2)j.
The boat is modelled as a particle.

(a)

Find the velocity of the boat at time t seconds.

(2)
(b)

Find the speed of the boat when t = 1.

(2)
(c)

Show that the acceleration of the boat is constant, and find its magnitude.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
differentiates both componentsM1
v = (6t − 2)i + (4 − 2t)jA1
Part (b)
v = 4i + 2j, uses PythagorasM1
speed = √20 = 4.47 m s−1A1
Part (c)
differentiates again: a = 6i − 2jM1
no t appears, so constant; |a| = √40 = 6.32 m s−2A1

Common error: Adding the components to get a speed (4 + 2 = 6). The components are perpendicular, so combine them with Pythagoras.

4IALM-M2-2-B-04 M2 · A2Easy6 marks

A robot lawnmower moves on a flat lawn. At time t seconds its velocity is v m s−1, where
v = 2ti + (6 − 3t)j.
When t = 0 the lawnmower is at the point with position vector (i − 4j) m relative to a fixed point O. The unit vectors i and j are directed due east and due north respectively. The lawnmower is modelled as a particle.

(a)

Find the position vector of the lawnmower at time t seconds.

(3)
(b)

Find the time at which the lawnmower is moving due east, and its position vector at that time.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
integrates both componentsM1
adds a constant vector and uses r = i − 4j when t = 0M1
r = (t2 + 1)i + (6t − 1.5t2 − 4)jA1
Part (b)
j-component zero: 6 − 3t = 0M1
t = 2, and the i-component, 4, is positive, so due eastA1
r = 5i + 2j ftA1

Exam tip: “Due east” needs two things: the north component of velocity is zero and the east component is positive. State both.

5IALM-M2-2-FB-01 M2 · A2Easy7 marks

A sledge is pulled in a straight line across level snow, starting from rest at time t = 0. Figure 1 shows the velocity–time graph of the sledge for 0 ≤ t ≤ 8, where v m s−1 is its velocity at time t seconds. The velocity is modelled by
v = pt + qt2,
where p and q are constants. The sledge is modelled as a particle.

01234567801234t / sv / m s⁻¹(2, 3)(8, 0)
(a)

Use the two marked points on Figure 1 to show that p = 2 and q = −0.25.

(3)
(b)

Find the deceleration of the sledge when t = 6.

(2)
(c)

The sledge reaches its greatest speed when t = 4. Find the distance it travels from t = 4 to t = 8.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
uses (2, 3): 2p + 4q = 3M1
uses (8, 0): 8p + 64q = 0M1
solves simultaneously: p = 2, q = −0.25A1
Part (b)
a = dv/dt = 2 − 0.5tM1
t = 6: a = −1, so the deceleration is 1 m s−2A1
Part (c)
∫ (2t − 0.25t2) dt = t2 − t3/12, with limits 4 and 8M1
(64 − 128/3) − (16 − 16/3) = 32/3 = 10.7 mA1

Exam tip: Each marked point on a graph gives one equation in the unknown constants. Two unknowns need two points: substitute both, then solve simultaneously.

6IALM-M2-2-FB-02 M2 · A2Easy6 marks

A shuttle moves along a straight horizontal rail. At time t seconds its displacement from a fixed point O on the rail is x metres, where
x = pt2 − 0.2t3,   0 ≤ t ≤ 6,
and p is a constant. Figure 1 shows the displacement–time graph of the shuttle. The shuttle is modelled as a particle.

012345601234567t / sx / m(6, 0)
(a)

Use the marked point on Figure 1 to show that p = 1.2.

(2)
(b)

Find the time, with 0 < t < 6, at which the shuttle is instantaneously at rest.

(2)
(c)

Find the total distance travelled by the shuttle for 0 ≤ t ≤ 6.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
uses x = 0 when t = 6: 36p − 43.2 = 0M1
p = 1.2A1
Part (b)
v = 2.4t − 0.6t2M1
0.6t(4 − t) = 0, so t = 4A1
Part (c)
at t = 4, x = 19.2 − 12.8 = 6.4 mM1
out 6.4 m and back to O: distance = 12.8 mA1

Exam tip: The graph ends on the t-axis, so the displacement at t = 6 is zero. The distance travelled is not zero: add the distance out to the turning point and the distance back.

7IALM-M2-2-FB-03 M2 · A2Easy7 marks

An ice skater moves on a horizontal rink. The unit vectors i and j are directed due east and due north respectively. At time t seconds the position vector of the skater relative to a fixed point O is r metres, where
r = pti + (t2 + qt + c)j,   0 ≤ t ≤ 4,
and p, q and c are constants. Figure 1 shows her path on a grid of 1 m squares, with her positions at t = 0, 1, 2, 3 and 4 marked. The skater is modelled as a particle.

123456789101112123456x / my / mOt = 0t = 1t = 2t = 3t = 4N
(a)

Use Figure 1 to find the values of p, q and c.

(3)
(b)

Find the velocity of the skater at time t seconds, and the time at which she is moving due east.

(2)
(c)

Find the speed of the skater when t = 4.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
t = 0 at (0, 5): c = 5B1
t = 2 at (6, 1): 2p = 6, so p = 3B1
4 + 2q + 5 = 1, so q = −4B1
Part (b)
v = 3i + (2t − 4)jM1
j-component zero at t = 2, where the i-component is 3 > 0A1
Part (c)
v = 3i + 4jM1
speed = √25 = 5 m s−1A1

Exam tip: A path diagram shows where the particle goes, not when. Match each marked point to its time label before substituting into r; any spare point is a free check on your constants.

8IALM-M2-2-FB-04 M2 · A2Easy7 marks

A remote-controlled car of mass 1.2 kg moves on a horizontal playground. The unit vectors i and j are directed due east and due north respectively. Figure 1 shows how the i-component and the j-component of the velocity of the car, in m s−1, vary with time t seconds for 0 ≤ t ≤ 4. Both graphs are straight lines. The car is modelled as a particle.

1234−2−101234567t / svelocity component / m s⁻¹east (i) componentnorth (j) component
(a)

Write down the velocity of the car at time t seconds, in terms of i, j and t.

(2)
(b)

Find the magnitude of the resultant force acting on the car.

(3)
(c)

Find the time at which the car is moving due east, and its speed at that time.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
i-component 2 + tB1
j-component 6 − 2t, so v = (2 + t)i + (6 − 2t)jB1
Part (b)
a = dv/dt = i − 2j (the gradients of the lines)M1
F = 1.2a = 1.2i − 2.4jM1
|F| = 1.2√5 = 2.68 NA1
Part (c)
6 − 2t = 0 gives t = 3M1
v = 5i, so the speed is 5 m s−1A1

Exam tip: The gradient of each component graph is the matching component of the acceleration. Straight lines mean the acceleration, and so the resultant force, is constant.

9IALM-M2-2-B-05 M2 · A2Medium8 marks

A motorboat moves in a straight line across a calm lake. Its engine is cut at time t = 0. For 0 ≤ t ≤ 10 its velocity is v m s−1, where
v = 8 − 1.6t + 0.08t2.
The boat is modelled as a particle.

(a)

Find the magnitude of the deceleration of the boat when t = 0.

(2)
(b)

Find the time taken for the speed of the boat to halve.

(2)
(c)

Find the distance travelled by the boat in the first 5 seconds after the engine is cut.

(3)
(d)

Given that v = 0.08(10 − t)2, show that the boat travels a total distance of 80/3 m between t = 0 and t = 10.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
a = dv/dt = −1.6 + 0.16tM1
deceleration = 1.6 m s−2A1
Part (b)
sets v = 4 and solves: t2 − 20t + 50 = 0M1
t = 10 − 5√2 = 2.93 s (10 + 5√2 is outside 0 ≤ t ≤ 10)A1
Part (c)
integrates v, at least one power of t increased by oneM1
x = 8t − 0.8t2 + 0.08t3/3, with limits 0 and 5M1
40 − 20 + 10/3 = 70/3 = 23.3 mA1
Part (d)
v ≥ 0 throughout, so distance = x(10) = 80 − 80 + 80/3 = 80/3 mB1

Exam tip: Since v = 0.08(10 − t)2 ≥ 0, the boat never turns round, so the definite integral of v is the distance travelled. Say so in one line.

10IALM-M2-2-B-06 M2 · A2Medium8 marks

A test platform of mass 12 kg moves along a vertical guide rail. At time t seconds its displacement above its starting level is y metres, where
y = 0.1t3 − 0.9t2 + 2.4t,   0 ≤ t ≤ 5.
The platform is modelled as a particle moving in a vertical line.

(a)

Show that the acceleration of the platform at time t seconds is 0.6(t − 3) m s−2.

(2)
(b)

Find the first time at which the platform is instantaneously at rest.

(2)
(c)

Find the greatest speed of the platform for 0 ≤ t ≤ 5.

(2)
(d)

Find the greatest magnitude of the resultant force on the platform for 0 ≤ t ≤ 5.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
v = 0.3t2 − 1.8t + 2.4M1
a = 0.6t − 1.8 = 0.6(t − 3)A1
Part (b)
0.3(t − 2)(t − 4) = 0 or equivalentM1
t = 2 sA1
Part (c)
v at t = 3 is −0.3, and compares |v| with the end values 2.4 (t = 0) and 0.9 (t = 5)M1
greatest speed = 2.4 m s−1A1
Part (d)
|a| is greatest at an end: 1.8 at t = 0, 1.2 at t = 5M1
F = 12 × 1.8 = 21.6 NA1

Common error: Taking the stationary point as the answer to every “greatest” question. Here the acceleration is linear in t, so its largest size is at an end of the interval.

11IALM-M2-2-B-07 M2 · A2Medium9 marks

A toy train moves along a straight track. It leaves a station O at time t = 0. At time t seconds its velocity is v m s−1, where
v = 0.5t2 − 3t + 4,   t ≥ 0.
The train is modelled as a particle.

(a)

Find the values of t at which the train is instantaneously at rest.

(2)
(b)

Find the least value of v, and interpret it.

(3)
(c)

Find the total distance travelled by the train in the first 5 seconds.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
0.5(t − 2)(t − 4) = 0 or equivalentM1
t = 2 and t = 4A1
Part (b)
dv/dt = t − 3 = 0M1
t = 3A1
v = −0.5: the train is moving back towards O at 0.5 m s−1A1
Part (c)
x = t3/6 − 1.5t2 + 4tM1
at t = 2, 4 and 5: x = 10/3, 8/3 and 10/3A1
splits the motion at the times of restM1
10/3 + 2/3 + 2/3 = 14/3 = 4.67 mA1

Exam tip: The displacement at t = 5 is 10/3 m, the same as at t = 2 — but the train has travelled further. Distance and displacement only agree if the particle never turns round.

12IALM-M2-2-B-08 M2 · A2Medium10 marks

A remote-controlled boat of mass 3 kg moves on a lake. At time t seconds its velocity is v m s−1, where
v = (t2 − 6t + 8)i + (2t + 1)j,   t ≥ 0.
The boat starts from a fixed point O at t = 0. The unit vectors i and j are directed due east and due north respectively. The boat is modelled as a particle.

(a)

Find the two times at which the boat is moving due north.

(3)
(b)

Find the magnitude of the resultant force on the boat at the later of these times.

(4)
(c)

Find the position vector of the boat relative to O at the later of these times.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
sets the i-component equal to zeroM1
(t − 2)(t − 4) = 0, so t = 2 and t = 4A1
j-component is 5 and 9 respectively, both positive, so both are due northB1
Part (b)
differentiates: a = (2t − 6)i + 2jM1
t = 4: a = 2i + 2jA1
F = 3a = 6i + 6jM1
|F| = 6√2 = 8.49 NA1
Part (c)
integrates both componentsM1
r = (t3/3 − 3t2 + 8t)i + (t2 + t)j, constant zeroA1
t = 4: r = (16/3)i + 20jA1

Exam tip: When asked for “the force”, give a vector; when asked for “the magnitude of the force”, give one positive number from Pythagoras.

13IALM-M2-2-B-09 M2 · A2Medium10 marks

A delivery drone of mass 1.5 kg flies in a horizontal plane. At time t seconds its position vector relative to a fixed point O is r metres, where
r = (t3 − 3t)i + (t2 + 2t)j,   t ≥ 0.
The drone is modelled as a particle.

(a)

Find the velocity of the drone at time t seconds.

(2)
(b)

Find the value of t at which the drone is moving parallel to the vector (3i + 2j).

(3)
(c)

Find the magnitude of the resultant force on the drone at this time.

(3)
(d)

Find the distance of the drone from O at this time.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
differentiates both componentsM1
v = (3t2 − 3)i + (2t + 2)jA1
Part (b)
2(3t2 − 3) = 3(2t + 2)M1
t2 − t − 2 = 0, so (t − 2)(t + 1) = 0M1
t = 2 (t = −1 rejected); check: v = 9i + 6j = 3(3i + 2j)A1
Part (c)
a = 6ti + 2jM1
t = 2: F = 1.5(12i + 2j) = 18i + 3jA1
|F| = √333 = 18.2 NA1
Part (d)
r = 2i + 8jM1
distance = √68 = 8.25 mA1

Exam tip: For “parallel to 3i + 2j”, the components of v must be in the ratio 3 : 2. Cross-multiplying, 2 × (i-component) = 3 × (j-component), avoids fractions.

14IALM-M2-2-B-10 M2 · A2Medium8 marks

A cyclist rides along a straight horizontal road. At time t = 0 she passes a point O with speed 2 m s−1. For 0 ≤ t ≤ 8 her acceleration is (1.2 − 0.15t) m s−2 in the direction of motion. The cyclist is modelled as a particle.

(a)

Find an expression for her velocity at time t seconds, for 0 ≤ t ≤ 8.

(3)
(b)

Find her speed when t = 8, and explain why this is her greatest speed for 0 ≤ t ≤ 8.

(2)
(c)

Find the distance she travels in the first 8 seconds.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
integrates the accelerationM1
includes a constant and uses v = 2 when t = 0M1
v = 2 + 1.2t − 0.075t2A1
Part (b)
v = 2 + 9.6 − 4.8 = 6.8 m s−1 ftB1
1.2 − 0.15t ≥ 0 for 0 ≤ t ≤ 8, so her speed never decreases in this intervalB1
Part (c)
integrates vM1
x = 2t + 0.6t2 − 0.025t3A1
t = 8: 16 + 38.4 − 12.8 = 41.6 mA1

Common error: Using s = ut + ½at2 with an average acceleration. The acceleration depends on t, so the constant-acceleration formulae do not apply.

15IALM-M2-2-FB-05 M2 · A2Medium8 marks

A particle P moves along a straight line. At time t seconds its displacement from a fixed point O on the line is x metres, where
x = pt2 − 2t5/2,   t ≥ 0,
and p is a constant. Figure 1 shows the displacement–time graph of P from t = 0 until P returns to O.

01234567024681012141618t / sx / m(1, 3)
(a)

Use the marked point on Figure 1 to show that p = 5.

(2)
(b)

Find the value of t, with t > 0, at which P is instantaneously at rest.

(2)
(c)

Find the acceleration of P at this instant.

(2)
(d)

Find the speed of P at the instant when it returns to O.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
substitutes t = 1, x = 3: p − 2 = 3M1
p = 5A1
Part (b)
v = dx/dt = 10t − 5t3/2M1
5t(2 − t1/2) = 0, so t = 4A1
Part (c)
a = dv/dt = 10 − 7.5t1/2M1
t = 4: a = 10 − 15 = −5 m s−2 (magnitude 5 m s−2, towards O)A1
Part (d)
t2(5 − 2t1/2) = 0 with t > 0 gives t1/2 = 2.5, t = 6.25, and substitutes into vM1
v = 62.5 − 78.125 = −15.625, so the speed is 15.6 m s−1 (accept 15.625)A1

Exam tip: Differentiate a fractional power exactly like any other power: d/dt(t5/2) = (5/2)t3/2. To solve v = 0, take out the common factor t and then solve for t1/2.

16IALM-M2-2-FB-06 M2 · A2Medium9 marks

A cart moves along a straight horizontal track in a mine and passes a fixed point O on the track at time t = 0. Figure 1 shows the velocity–time graph of the cart for 0 ≤ t ≤ 7, where v m s−1 is its velocity at time t seconds. The graph is a parabola, and v is a quadratic function of t. The cart is modelled as a particle.

1234567−4−20246810t / sv / m s⁻¹(0, 9)(2, 0)(6, 0)
(a)

Use the marked points on Figure 1 to show that v = 0.75t2 − 6t + 9.

(3)
(b)

Find the acceleration of the cart when t = 7.

(2)
(c)

Find the total distance travelled by the cart for 0 ≤ t ≤ 7.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
roots at t = 2 and t = 6: v = k(t − 2)(t − 6)M1
uses (0, 9): 12k = 9M1
k = 0.75, so v = 0.75t2 − 6t + 9A1
Part (b)
a = 1.5t − 6M1
t = 7: a = 4.5 m s−2A1
Part (c)
x = 0.25t3 − 3t2 + 9t, with x = 0 at t = 0M1
at t = 2, x = 8; at t = 6, x = 0A1
at t = 7, x = 1.75; splits the motion at t = 2 and t = 6M1
8 + 8 + 1.75 = 17.75 mA1

Common error: Integrating from 0 to 7 in one step gives the displacement, 1.75 m, not the distance. The graph crosses the t-axis twice, so split the integral where v = 0.

17IALM-M2-2-FB-07 M2 · A2Medium8 marks

In a shuttle-run drill, a runner starts from rest at time t = 0 and runs in a straight line until she comes to rest at the turning line. At time t seconds her velocity is v m s−1, where
v = pt − qt3/2,
and p and q are positive constants. Figure 1 shows the velocity–time graph of the runner. The curve passes through the marked point (1, 4), and the tangent to the curve at this point passes through the marked point (3, 10). The runner is modelled as a particle.

0123456789024681012t / sv / m s⁻¹(1, 4)(3, 10)tangent at (1, 4)
(a)

Use Figure 1 to show that p = 6 and q = 2.

(3)
(b)

Find the greatest speed of the runner.

(2)
(c)

The runner comes to rest at the turning line when t = 9. Find the distance from her starting point to the turning line.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
(1, 4) on the curve: p − q = 4B1
dv/dt = p − 1.5qt1/2, and the gradient of the tangent is (10 − 4)/(3 − 1) = 3, so p − 1.5q = 3M1
solves: 0.5q = 1, so q = 2 and p = 6A1
Part (b)
6 − 3t1/2 = 0, so t = 4M1
v = 24 − 16 = 8 m s−1A1
Part (c)
∫ (6t − 2t3/2) dt = 3t2 − 0.8t5/2M1
uses limits 0 and 9 (or x = 0 at t = 0)M1
243 − 0.8 × 243 = 48.6 mA1

Exam tip: The gradient of a tangent to a velocity–time graph is the acceleration at the point of contact. Find it from two points on the tangent, then set it equal to dv/dt at that time.

18IALM-M2-2-FB-08 M2 · A2Medium9 marks

A lift is raised vertically by a cable, starting from rest at ground level at time t = 0. Figure 1 shows the velocity–time graph of the lift for 0 ≤ t ≤ 8, where v m s−1 is its upward velocity at time t seconds. In the model,
v = kt2 for 0 ≤ t ≤ 2,   v is constant for 2 ≤ t ≤ 6,   v = 3 − k(t − 6)2 for 6 ≤ t ≤ 8,
where k is a constant. The lift and its passengers have total mass 800 kg and are modelled as a particle. Take g = 9.8 m s−2.

01234567801234t / sv / m s⁻¹(2, 3)(6, 3)(8, 0)
(a)

Use Figure 1 to find the value of k.

(2)
(b)

Find the height risen by the lift in the 8 seconds.

(4)
(c)

Find the tension in the cable when t = 1.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
v = 3 when t = 2: 4k = 3M1
k = 0.75A1
Part (b)
integrates 0.75t2: 0.25t3M1
first 2 s: 2 mA1
2 ≤ t ≤ 6: 3 × 4 = 12 mB1
6 ≤ t ≤ 8: ∫ (3 − 0.75(t − 6)2) dt = 6 − 2 = 4 m, total 18 mA1
Part (c)
a = 2kt = 1.5 m s−2 at t = 1M1
T − 800g = 800 × 1.5M1
T = 800 × 11.3 = 9040 NA1

Common error: Writing T = ma for the lift. The resultant upward force is T − mg, so the weight must appear in the equation of motion.

19IALM-M2-2-FB-09 M2 · A2Medium9 marks

A kayak moves on a lake. The unit vectors i and j are directed due east and due north respectively. Figure 1 is a map of part of the lake, with grid squares of side 1 m and a fixed point O. The kayak starts from the point A at time t = 0, and its path passes a buoy at the point B. At time t seconds the velocity of the kayak is v m s−1, where
v = (6 − 2t)i + (2t − 2)j.
The kayak and its paddler have total mass 90 kg and are modelled as a particle.

−5−4−3−2−1123456−112345678910x / my / mOAB (buoy)N
(a)

Find the position vector of the kayak, relative to O, at time t seconds.

(3)
(b)

Find the value of t when the kayak passes the buoy.

(3)
(c)

Find the magnitude of the resultant force acting on the kayak and its paddler.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
integrates both componentsM1
uses r = −4i + j when t = 0, read from Figure 1M1
r = (6t − t2 − 4)i + (t2 − 2t + 1)jA1
Part (b)
i-component 4: t2 − 6t + 8 = 0, so t = 2 or t = 4M1
j-component 9: (t − 1)2 = 9, so t = 4 (as t ≥ 0)M1
t = 4 is the only common valueA1
Part (c)
a = dv/dt = −2i + 2jM1
F = 90a = −180i + 180jM1
|F| = 180√2 = 255 NA1

Exam tip: To show a particle passes through a point, both components must give the same value of t. Here the i-component is 4 at t = 2 as well, but the kayak is then at (4, 1), not at the buoy.

20IALM-M2-2-FB-10 M2 · A2Medium9 marks

A kite-surfer moves on the surface of the sea, starting from a fixed point O at time t = 0. The unit vectors i and j are directed due east and due north respectively. At time t seconds the position vector of the kite-surfer relative to O is r metres, where
r = (t2 + 2t)i + (6t − t2)j,   0 ≤ t ≤ 5.
Figure 1 shows the path on a grid of 1 m squares, together with an arrow, drawn between two grid points, showing the direction in which the wind is blowing. The kite-surfer is modelled as a particle.

4812162024283236246810x / my / mOwind directionN
(a)

Find the velocity of the kite-surfer at time t seconds.

(2)
(b)

Use Figure 1 to find the time at which the kite-surfer is moving in the direction in which the wind is blowing.

(3)
(c)

Find the distance of the kite-surfer from O at this time.

(2)
(d)

Find the position vector of the kite-surfer at the instant when it is moving due east.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
differentiates both componentsM1
v = (2t + 2)i + (6 − 2t)jA1
Part (b)
arrow from (21, 1) to (30, 4): direction 9i + 3j, parallel to 3i + jB1
2t + 2 = 3(6 − 2t)M1
t = 2, where v = 6i + 2j = 2(3i + j), the same directionA1
Part (c)
r = 8i + 8jM1
distance = 8√2 = 11.3 mA1
Part (d)
6 − 2t = 0, so t = 3, where the i-component is 8 > 0M1
r = 15i + 9jA1

Common error: Using the position of the arrow's tip, 30i + 4j, as the wind direction. A direction is the difference between the two ends of the arrow: (30 − 21)i + (4 − 1)j.

21IALM-M2-2-B-11 M2 · A2Hard14 marks

A bead is threaded on a straight horizontal wire. At time t = 0 the bead is at a fixed point O on the wire. At time t seconds its velocity is v m s−1, where
v = 2t2 − 8t + 6   for 0 ≤ t ≤ 3,
v = 12 − 108t−2   for t > 3.
The bead is modelled as a particle.

(a)

Find the values of t, for 0 ≤ t ≤ 3, at which the bead is instantaneously at rest.

(2)
(b)

Find the greatest speed of the bead for 0 ≤ t ≤ 3.

(3)
(c)

Find the total distance travelled by the bead for 0 ≤ t ≤ 3.

(4)
(d)

Find the acceleration of the bead when t = 6.

(2)
(e)

Find the distance travelled by the bead between t = 3 and t = 6.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
2(t − 1)(t − 3) = 0 or equivalentM1
t = 1 and t = 3A1
Part (b)
dv/dt = 4t − 8 = 0 gives t = 2, v = −2M1
compares with the end values v = 6 at t = 0 and v = 0 at t = 3M1
greatest speed = 6 m s−1, at t = 0A1
Part (c)
x = 2t3/3 − 4t2 + 6tM1
at t = 1, x = 8/3A1
at t = 3, x = 0A1
distance = 8/3 + 8/3 = 16/3 m (5.33 m)A1
Part (d)
a = 216t−3M1
t = 6: a = 1 m s−2A1
Part (e)
for t > 3, 108t−2 < 12 so v > 0: no change of directionB1
∫ (12 − 108t−2) dt = 12t + 108t−1, with limits 3 and 6M1
(72 + 18) − (36 + 36) = 18 mA1

Exam tip: “Greatest speed” over an interval: compare |v| at every stationary point and at both ends. Here the turning point gives only 2 m s−1; the start gives 6 m s−1.

22IALM-M2-2-B-12 M2 · A2Hard13 marks

A test sled moves along a straight horizontal rail. It starts from rest at a fixed point O on the rail at time t = 0. At time t seconds its acceleration is (pt + q) m s−2, where p and q are constants. When t = 2 the velocity of the sled is 8 m s−1, and when t = 4 the sled is again instantaneously at rest. The sled is modelled as a particle.

(a)

Find the values of p and q.

(5)
(b)

Find the displacement of the sled from O when t = 4.

(3)
(c)

Find the time at which the sled returns to O.

(3)
(d)

Find the speed of the sled as it returns to O.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
integrates: v = ½pt2 + qt, constant zero as it starts from restM1
t = 2: 2p + 2q = 8M1
t = 4: 8p + 4q = 0M1
solves the simultaneous equationsM1
p = −4, q = 8A1
Part (b)
v = 8t − 2t2, integratesM1
x = 4t2 − 2t3/3 ftA1
t = 4: x = 64 − 128/3 = 64/3 m (21.3 m)A1
Part (c)
sets x = 0M1
t2(4 − 2t/3) = 0M1
t = 6 (t = 0 is the start)A1
Part (d)
t = 6: v = 48 − 72 = −24M1
speed = 24 m s−1A1

Exam tip: Unknown constants in an acceleration are found exactly like constants of integration: integrate first, then substitute each piece of information to get one equation per unknown.

23IALM-M2-2-B-13 M2 · A2Hard12 marks

A small hovercraft of mass 0.5 kg moves on a large flat sheet of ice. At time t seconds the resultant force on it is F newtons, where
F = (3t − 2)i − j.
When t = 0 the hovercraft has velocity (2i + 6j) m s−1 and is at the point with position vector (−i + 3j) m relative to a fixed point O. The unit vectors i and j are directed due east and due north respectively. The hovercraft is modelled as a particle.

(a)

Find the velocity of the hovercraft at time t seconds.

(3)
(b)

Show that the hovercraft is never moving due north or due south.

(2)
(c)

Find the speed of the hovercraft at the instant when it is moving due east.

(2)
(d)

Find the position vector of the hovercraft at time t seconds.

(3)
(e)

Find the distance of the hovercraft from O at the instant when it is moving due east.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
a = F/0.5 = (6t − 4)i − 2jM1
integrates and uses the initial velocityM1
v = (3t2 − 4t + 2)i + (6 − 2t)jA1
Part (b)
needs the i-component zero: 3t2 − 4t + 2 = 0, discriminant 16 − 24 = −8M1
negative, so no real t: the i-component is never zeroA1
Part (c)
j-component zero at t = 3, where the i-component is 27 − 12 + 2 = 17 > 0M1
speed = 17 m s−1A1
Part (d)
integrates vM1
uses r = −i + 3j when t = 0M1
r = (t3 − 2t2 + 2t − 1)i + (3 + 6t − t2)jA1
Part (e)
t = 3: r = 14i + 12jM1
distance = √340 = 18.4 mA1

Common error: Integrating the force instead of the acceleration. Divide F by the mass first; here that doubles every component.

24IALM-M2-2-B-14 M2 · A2Hard13 marks

A jet-ski of mass 250 kg moves on the sea. At time t seconds its velocity is v m s−1, where
v = (t2 − 5t + 6)i + (t2 − 4)j,   t ≥ 0.
The unit vectors i and j are directed due east and due north respectively. The jet-ski is modelled as a particle.

(a)

Show that the jet-ski is instantaneously at rest at exactly one time, and state this time.

(3)
(b)

Find the time at which the jet-ski is moving due north. Give a reason for rejecting any other value of t.

(3)
(c)

Find the value of t, other than t = 2, at which the jet-ski is moving parallel to (i − j), and find its speed at that time.

(4)
(d)

Find the magnitude of the resultant force on the jet-ski when it is moving due north.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
i-component zero when (t − 2)(t − 3) = 0M1
j-component zero when t = 2 (as t ≥ 0)M1
both are zero only at t = 2A1
Part (b)
i-component zero at t = 2 or t = 3M1
at t = 2 the jet-ski is at rest, so it is not moving in any directionB1
t = 3, where the j-component is 5 > 0A1
Part (c)
t2 − 5t + 6 = −(t2 − 4)M1
2t2 − 5t + 2 = 0, so t = 0.5 or t = 2A1
t = 0.5: v = 3.75i − 3.75jA1
speed = 3.75√2 = 5.30 m s−1A1
Part (d)
a = (2t − 5)i + 2tjM1
t = 3: a = i + 6j, so F = 250i + 1500jA1
|F| = 250√37 = 1520 NA1

Exam tip: A particle at rest has no direction of motion. Whenever a “due north” or “parallel to” equation has a root where the velocity is the zero vector, reject that root and say why.

25IALM-M2-2-FB-11 M2 · A2Hard12 marks

A metro train travels along a straight horizontal track between two stations. It starts from rest at the first station at time t = 0 and comes to rest at the second station. At time t seconds its velocity is v m s−1, where
v = kt2(60 − t),
and k is a constant. Figure 1 shows the velocity–time graph of the train, with two marked points. The train has mass 1.2 × 105 kg and is modelled as a particle.

010203040506004812162024t / sv / m s⁻¹(40, 20)(60, 0)
(a)

Show, by differentiation, that the greatest velocity occurs when t = 40, and use Figure 1 to show that k = 1/1600.

(4)
(b)

Find the distance between the two stations.

(3)
(c)

Find the greatest magnitude of the acceleration of the train for 0 ≤ t ≤ 60.

(3)
(d)

Find the greatest magnitude of the resultant force on the train.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
dv/dt = k(120t − 3t2)M1
3kt(40 − t) = 0, so t = 40 for 0 < t < 60A1
greatest velocity 20 from Figure 1: k × 1600 × 20 = 20M1
k = 1/1600A1
Part (b)
integrates: x = k(20t3 − t4/4)M1
uses limits 0 and 60M1
(4 320 000 − 3 240 000)/1600 = 675 mA1
Part (c)
a = (120t − 3t2)/1600; da/dt = 0 at t = 20, where a = 0.75M1
compares with the ends: a = 0 at t = 0 and a = −2.25 at t = 60M1
greatest magnitude 2.25 m s−2, as the train arrivesA1
Part (d)
F = 1.2 × 105 × 2.25M1
= 2.7 × 105 NA1

Exam tip: For the greatest size of a quantity on an interval, check its stationary points and both end values, and compare magnitudes. Here the end value t = 60 wins.

26IALM-M2-2-FB-12 M2 · A2Hard14 marks

A small inspection probe moves along a straight horizontal pipe. At time t = 0 it passes a fixed point O in the pipe with speed 5 m s−1 in the negative direction. At time t seconds its velocity is v m s−1 and its displacement from O is x metres. Figure 1 shows the acceleration–time graph of the probe for 0 ≤ t ≤ 6, which is a straight line. The probe is modelled as a particle.

123456−6−4−20246t / sa / m s⁻²
(a)

Use Figure 1 to find an expression for the acceleration of the probe at time t seconds.

(2)
(b)

Show that v = −t2 + 6t − 5.

(2)
(c)

Find the times at which the probe is instantaneously at rest.

(2)
(d)

Find the greatest speed of the probe for 0 ≤ t ≤ 6.

(3)
(e)

Find the total distance travelled by the probe for 0 ≤ t ≤ 6.

(5)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient −2 and intercept 6M1
a = 6 − 2tA1
Part (b)
integrates: v = 6t − t2 + CM1
v = −5 when t = 0, so C = −5A1
Part (c)
−(t − 1)(t − 5) = 0M1
t = 1 and t = 5A1
Part (d)
a = 0 at t = 3, where v = 4M1
end values: v = −5 at t = 0 and at t = 6M1
greatest speed = 5 m s−1, at both endsA1
Part (e)
x = −t3/3 + 3t2 − 5tM1
at t = 1, x = −7/3A1
at t = 5, x = 25/3; at t = 6, x = 6A1
splits at the times of rest: 7/3 + 32/3 + 7/3M1
= 46/3 = 15.3 mA1

Common error: Taking the stationary value of v (where a = 0) as the greatest speed. Speed is |v|: here the ends give 5 m s−1, more than the 4 m s−1 at t = 3.

27IALM-M2-2-FB-13 M2 · A2Hard13 marks

Two model cars, P and Q, move on a horizontal car park. The unit vectors i and j are directed due east and due north respectively. Figure 1 is a plan with grid squares of side 1 m and a fixed point O. It shows the starting points, A of P and B of Q, at time t = 0, and the paths of the cars for 0 ≤ t ≤ 2.3. At time t seconds the velocities of P and Q are
vP = (3t2 − 2)i + 4tj m s−1   and   vQ = −2ti − j m s−1.
The cars are modelled as particles.

1234567891011−4−3−2−112345678x / my / mOABpath of Qpath of PN
(a)

Find the position vector of P, relative to O, at time t seconds.

(3)
(b)

Find the position vector of Q, relative to O, at time t seconds.

(2)
(c)

Show that the cars collide, and find the time of the collision and the position vector of the point where it happens.

(4)
(d)

Find the speed of each car at the instant of the collision.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
integrates both componentsM1
adds 2i − 3j, the position of A from Figure 1M1
rP = (t3 − 2t + 2)i + (2t2 − 3)jA1
Part (b)
integrates and adds 10i + 7jM1
rQ = (10 − t2)i + (7 − t)jA1
Part (c)
j-components equal: 2t2 + t − 10 = 0, (2t + 5)(t − 2) = 0, so t = 2M1
i-components equal: t3 + t2 − 2t − 8 = 0M1
t = 2 satisfies this too: 8 + 4 − 4 − 8 = 0, so both cars are at the same point at the same timeA1
collision at t = 2, at 6i + 5jA1
Part (d)
vP = 10i + 8jM1
speed of P = √164 = 12.8 m s−1A1
vQ = −4i − jM1
speed of Q = √17 = 4.12 m s−1A1

Exam tip: Paths that cross on a diagram do not prove a collision. The particles must be at the same point at the same time, so both components must agree for one value of t.

28IALM-M2-2-FB-14 M2 · A2Hard13 marks

A particle of mass 2 kg moves on a smooth horizontal plane. The unit vectors i and j are directed due east and due north respectively. At time t seconds the acceleration of the particle is a m s−2, where
a = pti + qj,
and p and q are constants. Figure 1 is a velocity diagram, drawn on a grid of unit squares, showing the velocity of the particle at t = 0 and at t = 2. The particle is at a fixed point O when t = 0.

−4−3−2−112−112345i-component / m s⁻¹j-component / m s⁻¹Ovelocity at t = 0velocity at t = 2
(a)

Use Figure 1 to find the values of p and q.

(4)
(b)

Find the exact value of t at which the particle is moving due north.

(3)
(c)

Find the magnitude of the resultant force on the particle at this time.

(3)
(d)

Find the position vector of the particle at the instant, after t = 0, when it is due north of O.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
v = (½pt2 + c1)i + (qt + c2)jM1
uses v = −3i + 4j when t = 0M1
uses v = i when t = 2: 2p − 3 = 1, so p = 2A1
2q + 4 = 0, so q = −2A1
Part (b)
i-component t2 − 3 = 0M1
t = √3A1
j-component 4 − 2√3 > 0, so the motion is north, not southB1
Part (c)
a = 2√3i − 2jM1
F = 2a = 4√3i − 4jM1
|F| = √(48 + 16) = 8 NA1
Part (d)
r = (t3/3 − 3t)i + (4t − t2)jM1
t3/3 − 3t = 0 with t > 0, so t = 3M1
r = 3j (north of O, as 3 > 0)A1

Common error: Treating the arrows as accelerations. Both arrows are velocities; the change from one to the other is the integral of a from t = 0 to t = 2.