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-01M2 · 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)
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Answer
Mark
Notes
Part (a)
differentiates, at least one power of t reduced by one
M1
v = 1.5t2 − 12t + 18
A1
Part (b)
sets v = 0 and solves: 1.5(t − 2)(t − 6) = 0
M1
t = 2 and t = 6
A1
Part (c)
substitutes the smaller value, t = 2, into x
M1
x = 4 − 24 + 36 = 16 m
A1
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.
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)
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Answer
Mark
Notes
Part (a)
a = dv/dt = 2.4 − 0.6t
M1
t = 2: a = 1.2 m s−2
A1
Part (b)
a = 0 gives t = 4
M1
v = 9.6 − 4.8 = 4.8 m s−1
A1
Part (c)
integrates v, at least one power of t increased by one
M1
h = 1.2t2 − 0.1t3, constant zero since h = 0 when t = 0
A1
h = 76.8 − 51.2 = 25.6 m
A1
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).
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)
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Answer
Mark
Notes
Part (a)
differentiates both components
M1
v = (6t − 2)i + (4 − 2t)j
A1
Part (b)
v = 4i + 2j, uses Pythagoras
M1
speed = √20 = 4.47 m s−1
A1
Part (c)
differentiates again: a = 6i − 2j
M1
no t appears, so constant; |a| = √40 = 6.32 m s−2
A1
Common error: Adding the components to get a speed (4 + 2 = 6). The components are perpendicular, so combine them with Pythagoras.
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)
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Answer
Mark
Notes
Part (a)
integrates both components
M1
adds a constant vector and uses r = i − 4j when t = 0
M1
r = (t2 + 1)i + (6t − 1.5t2 − 4)j
A1
Part (b)
j-component zero: 6 − 3t = 0
M1
t = 2, and the i-component, 4, is positive, so due east
A1
r = 5i + 2j ft
A1
Exam tip: “Due east” needs two things: the north component of velocity is zero and the east component is positive. State both.
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.
(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)
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Answer
Mark
Notes
Part (a)
uses (2, 3): 2p + 4q = 3
M1
uses (8, 0): 8p + 64q = 0
M1
solves simultaneously: p = 2, q = −0.25
A1
Part (b)
a = dv/dt = 2 − 0.5t
M1
t = 6: a = −1, so the deceleration is 1 m s−2
A1
Part (c)
∫ (2t − 0.25t2) dt = t2 − t3/12, with limits 4 and 8
M1
(64 − 128/3) − (16 − 16/3) = 32/3 = 10.7 m
A1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
uses x = 0 when t = 6: 36p − 43.2 = 0
M1
p = 1.2
A1
Part (b)
v = 2.4t − 0.6t2
M1
0.6t(4 − t) = 0, so t = 4
A1
Part (c)
at t = 4, x = 19.2 − 12.8 = 6.4 m
M1
out 6.4 m and back to O: distance = 12.8 m
A1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
t = 0 at (0, 5): c = 5
B1
t = 2 at (6, 1): 2p = 6, so p = 3
B1
4 + 2q + 5 = 1, so q = −4
B1
Part (b)
v = 3i + (2t − 4)j
M1
j-component zero at t = 2, where the i-component is 3 > 0
A1
Part (c)
v = 3i + 4j
M1
speed = √25 = 5 m s−1
A1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
i-component 2 + t
B1
j-component 6 − 2t, so v = (2 + t)i + (6 − 2t)j
B1
Part (b)
a = dv/dt = i − 2j (the gradients of the lines)
M1
F = 1.2a = 1.2i − 2.4j
M1
|F| = 1.2√5 = 2.68 N
A1
Part (c)
6 − 2t = 0 gives t = 3
M1
v = 5i, so the speed is 5 m s−1
A1
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.
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)
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Answer
Mark
Notes
Part (a)
a = dv/dt = −1.6 + 0.16t
M1
deceleration = 1.6 m s−2
A1
Part (b)
sets v = 4 and solves: t2 − 20t + 50 = 0
M1
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 one
M1
x = 8t − 0.8t2 + 0.08t3/3, with limits 0 and 5
M1
40 − 20 + 10/3 = 70/3 = 23.3 m
A1
Part (d)
v ≥ 0 throughout, so distance = x(10) = 80 − 80 + 80/3 = 80/3 m
B1
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.
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)
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Answer
Mark
Notes
Part (a)
v = 0.3t2 − 1.8t + 2.4
M1
a = 0.6t − 1.8 = 0.6(t − 3)
A1
Part (b)
0.3(t − 2)(t − 4) = 0 or equivalent
M1
t = 2 s
A1
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−1
A1
Part (d)
|a| is greatest at an end: 1.8 at t = 0, 1.2 at t = 5
M1
F = 12 × 1.8 = 21.6 N
A1
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.
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)
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Answer
Mark
Notes
Part (a)
0.5(t − 2)(t − 4) = 0 or equivalent
M1
t = 2 and t = 4
A1
Part (b)
dv/dt = t − 3 = 0
M1
t = 3
A1
v = −0.5: the train is moving back towards O at 0.5 m s−1
A1
Part (c)
x = t3/6 − 1.5t2 + 4t
M1
at t = 2, 4 and 5: x = 10/3, 8/3 and 10/3
A1
splits the motion at the times of rest
M1
10/3 + 2/3 + 2/3 = 14/3 = 4.67 m
A1
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.
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)
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Answer
Mark
Notes
Part (a)
sets the i-component equal to zero
M1
(t − 2)(t − 4) = 0, so t = 2 and t = 4
A1
j-component is 5 and 9 respectively, both positive, so both are due north
B1
Part (b)
differentiates: a = (2t − 6)i + 2j
M1
t = 4: a = 2i + 2j
A1
F = 3a = 6i + 6j
M1
|F| = 6√2 = 8.49 N
A1
Part (c)
integrates both components
M1
r = (t3/3 − 3t2 + 8t)i + (t2 + t)j, constant zero
A1
t = 4: r = (16/3)i + 20j
A1
Exam tip: When asked for “the force”, give a vector; when asked for “the magnitude of the force”, give one positive number from Pythagoras.
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)
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Answer
Mark
Notes
Part (a)
differentiates both components
M1
v = (3t2 − 3)i + (2t + 2)j
A1
Part (b)
2(3t2 − 3) = 3(2t + 2)
M1
t2 − t − 2 = 0, so (t − 2)(t + 1) = 0
M1
t = 2 (t = −1 rejected); check: v = 9i + 6j = 3(3i + 2j)
A1
Part (c)
a = 6ti + 2j
M1
t = 2: F = 1.5(12i + 2j) = 18i + 3j
A1
|F| = √333 = 18.2 N
A1
Part (d)
r = 2i + 8j
M1
distance = √68 = 8.25 m
A1
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.
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)
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Answer
Mark
Notes
Part (a)
integrates the acceleration
M1
includes a constant and uses v = 2 when t = 0
M1
v = 2 + 1.2t − 0.075t2
A1
Part (b)
v = 2 + 9.6 − 4.8 = 6.8 m s−1 ft
B1
1.2 − 0.15t ≥ 0 for 0 ≤ t ≤ 8, so her speed never decreases in this interval
B1
Part (c)
integrates v
M1
x = 2t + 0.6t2 − 0.025t3
A1
t = 8: 16 + 38.4 − 12.8 = 41.6 m
A1
Common error: Using s = ut + ½at2 with an average acceleration. The acceleration depends on t, so the constant-acceleration formulae do not apply.
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.
(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)
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Answer
Mark
Notes
Part (a)
substitutes t = 1, x = 3: p − 2 = 3
M1
p = 5
A1
Part (b)
v = dx/dt = 10t − 5t3/2
M1
5t(2 − t1/2) = 0, so t = 4
A1
Part (c)
a = dv/dt = 10 − 7.5t1/2
M1
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 v
M1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
roots at t = 2 and t = 6: v = k(t − 2)(t − 6)
M1
uses (0, 9): 12k = 9
M1
k = 0.75, so v = 0.75t2 − 6t + 9
A1
Part (b)
a = 1.5t − 6
M1
t = 7: a = 4.5 m s−2
A1
Part (c)
x = 0.25t3 − 3t2 + 9t, with x = 0 at t = 0
M1
at t = 2, x = 8; at t = 6, x = 0
A1
at t = 7, x = 1.75; splits the motion at t = 2 and t = 6
M1
8 + 8 + 1.75 = 17.75 m
A1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
(1, 4) on the curve: p − q = 4
B1
dv/dt = p − 1.5qt1/2, and the gradient of the tangent is (10 − 4)/(3 − 1) = 3, so p − 1.5q = 3
M1
solves: 0.5q = 1, so q = 2 and p = 6
A1
Part (b)
6 − 3t1/2 = 0, so t = 4
M1
v = 24 − 16 = 8 m s−1
A1
Part (c)
∫ (6t − 2t3/2) dt = 3t2 − 0.8t5/2
M1
uses limits 0 and 9 (or x = 0 at t = 0)
M1
243 − 0.8 × 243 = 48.6 m
A1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
v = 3 when t = 2: 4k = 3
M1
k = 0.75
A1
Part (b)
integrates 0.75t2: 0.25t3
M1
first 2 s: 2 m
A1
2 ≤ t ≤ 6: 3 × 4 = 12 m
B1
6 ≤ t ≤ 8: ∫ (3 − 0.75(t − 6)2) dt = 6 − 2 = 4 m, total 18 m
A1
Part (c)
a = 2kt = 1.5 m s−2 at t = 1
M1
T − 800g = 800 × 1.5
M1
T = 800 × 11.3 = 9040 N
A1
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.
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.
(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)
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Answer
Mark
Notes
Part (a)
integrates both components
M1
uses r = −4i + j when t = 0, read from Figure 1
M1
r = (6t − t2 − 4)i + (t2 − 2t + 1)j
A1
Part (b)
i-component 4: t2 − 6t + 8 = 0, so t = 2 or t = 4
M1
j-component 9: (t − 1)2 = 9, so t = 4 (as t ≥ 0)
M1
t = 4 is the only common value
A1
Part (c)
a = dv/dt = −2i + 2j
M1
F = 90a = −180i + 180j
M1
|F| = 180√2 = 255 N
A1
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.
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.
(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
Answer
Mark
Notes
Part (a)
differentiates both components
M1
v = (2t + 2)i + (6 − 2t)j
A1
Part (b)
arrow from (21, 1) to (30, 4): direction 9i + 3j, parallel to 3i + j
B1
2t + 2 = 3(6 − 2t)
M1
t = 2, where v = 6i + 2j = 2(3i + j), the same direction
A1
Part (c)
r = 8i + 8j
M1
distance = 8√2 = 11.3 m
A1
Part (d)
6 − 2t = 0, so t = 3, where the i-component is 8 > 0
M1
r = 15i + 9j
A1
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.
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
Answer
Mark
Notes
Part (a)
2(t − 1)(t − 3) = 0 or equivalent
M1
t = 1 and t = 3
A1
Part (b)
dv/dt = 4t − 8 = 0 gives t = 2, v = −2
M1
compares with the end values v = 6 at t = 0 and v = 0 at t = 3
M1
greatest speed = 6 m s−1, at t = 0
A1
Part (c)
x = 2t3/3 − 4t2 + 6t
M1
at t = 1, x = 8/3
A1
at t = 3, x = 0
A1
distance = 8/3 + 8/3 = 16/3 m (5.33 m)
A1
Part (d)
a = 216t−3
M1
t = 6: a = 1 m s−2
A1
Part (e)
for t > 3, 108t−2 < 12 so v > 0: no change of direction
B1
∫ (12 − 108t−2) dt = 12t + 108t−1, with limits 3 and 6
M1
(72 + 18) − (36 + 36) = 18 m
A1
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.
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
Answer
Mark
Notes
Part (a)
integrates: v = ½pt2 + qt, constant zero as it starts from rest
M1
t = 2: 2p + 2q = 8
M1
t = 4: 8p + 4q = 0
M1
solves the simultaneous equations
M1
p = −4, q = 8
A1
Part (b)
v = 8t − 2t2, integrates
M1
x = 4t2 − 2t3/3 ft
A1
t = 4: x = 64 − 128/3 = 64/3 m (21.3 m)
A1
Part (c)
sets x = 0
M1
t2(4 − 2t/3) = 0
M1
t = 6 (t = 0 is the start)
A1
Part (d)
t = 6: v = 48 − 72 = −24
M1
speed = 24 m s−1
A1
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.
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.
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
Answer
Mark
Notes
Part (a)
i-component zero when (t − 2)(t − 3) = 0
M1
j-component zero when t = 2 (as t ≥ 0)
M1
both are zero only at t = 2
A1
Part (b)
i-component zero at t = 2 or t = 3
M1
at t = 2 the jet-ski is at rest, so it is not moving in any direction
B1
t = 3, where the j-component is 5 > 0
A1
Part (c)
t2 − 5t + 6 = −(t2 − 4)
M1
2t2 − 5t + 2 = 0, so t = 0.5 or t = 2
A1
t = 0.5: v = 3.75i − 3.75j
A1
speed = 3.75√2 = 5.30 m s−1
A1
Part (d)
a = (2t − 5)i + 2tj
M1
t = 3: a = i + 6j, so F = 250i + 1500j
A1
|F| = 250√37 = 1520 N
A1
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.
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.
(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
Answer
Mark
Notes
Part (a)
dv/dt = k(120t − 3t2)
M1
3kt(40 − t) = 0, so t = 40 for 0 < t < 60
A1
greatest velocity 20 from Figure 1: k × 1600 × 20 = 20
M1
k = 1/1600
A1
Part (b)
integrates: x = k(20t3 − t4/4)
M1
uses limits 0 and 60
M1
(4 320 000 − 3 240 000)/1600 = 675 m
A1
Part (c)
a = (120t − 3t2)/1600; da/dt = 0 at t = 20, where a = 0.75
M1
compares with the ends: a = 0 at t = 0 and a = −2.25 at t = 60
M1
greatest magnitude 2.25 m s−2, as the train arrives
A1
Part (d)
F = 1.2 × 105 × 2.25
M1
= 2.7 × 105 N
A1
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.
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.
(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
Answer
Mark
Notes
Part (a)
gradient −2 and intercept 6
M1
a = 6 − 2t
A1
Part (b)
integrates: v = 6t − t2 + C
M1
v = −5 when t = 0, so C = −5
A1
Part (c)
−(t − 1)(t − 5) = 0
M1
t = 1 and t = 5
A1
Part (d)
a = 0 at t = 3, where v = 4
M1
end values: v = −5 at t = 0 and at t = 6
M1
greatest speed = 5 m s−1, at both ends
A1
Part (e)
x = −t3/3 + 3t2 − 5t
M1
at t = 1, x = −7/3
A1
at t = 5, x = 25/3; at t = 6, x = 6
A1
splits at the times of rest: 7/3 + 32/3 + 7/3
M1
= 46/3 = 15.3 m
A1
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.
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.
(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
Answer
Mark
Notes
Part (a)
integrates both components
M1
adds 2i − 3j, the position of A from Figure 1
M1
rP = (t3 − 2t + 2)i + (2t2 − 3)j
A1
Part (b)
integrates and adds 10i + 7j
M1
rQ = (10 − t2)i + (7 − t)j
A1
Part (c)
j-components equal: 2t2 + t − 10 = 0, (2t + 5)(t − 2) = 0, so t = 2
M1
i-components equal: t3 + t2 − 2t − 8 = 0
M1
t = 2 satisfies this too: 8 + 4 − 4 − 8 = 0, so both cars are at the same point at the same time
A1
collision at t = 2, at 6i + 5j
A1
Part (d)
vP = 10i + 8j
M1
speed of P = √164 = 12.8 m s−1
A1
vQ = −4i − j
M1
speed of Q = √17 = 4.12 m s−1
A1
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.
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.
(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
Answer
Mark
Notes
Part (a)
v = (½pt2 + c1)i + (qt + c2)j
M1
uses v = −3i + 4j when t = 0
M1
uses v = i when t = 2: 2p − 3 = 1, so p = 2
A1
2q + 4 = 0, so q = −2
A1
Part (b)
i-component t2 − 3 = 0
M1
t = √3
A1
j-component 4 − 2√3 > 0, so the motion is north, not south
B1
Part (c)
a = 2√3i − 2j
M1
F = 2a = 4√3i − 4j
M1
|F| = √(48 + 16) = 8 N
A1
Part (d)
r = (t3/3 − 3t)i + (4t − t2)j
M1
t3/3 − 3t = 0 with t > 0, so t = 3
M1
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.
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