kinematics [285 marks] - github pages[2 marks] t=2 (exact), 0.667, t=4 3 4b.hence or otherwise, find...

Kinematics [285 marks]

1a.

A particle moves along a straight line so that its velocity, m s , after seconds is given by , for 0 ≤ ≤ 5.

Find when the particle is at rest.

Markschemevalid approach (M1)

eg , sketch of graph

2.95195

(exact), (s) A1 N2

[2 marks]

v −1 tv (t) = 1.4t − 2.7 t

v (t) = 0

t = log1.42.7 t = 2.95

1b. Find the acceleration of the particle when .


eg ,

0.659485

= 1.96 ln 1.4 (exact), = 0.659 (m s ) A1 N2

[2 marks]

t = 2

a (t) = v′ (t) v′ (2)

a (2) a (2) −2

1c. Find the total distance travelled by the particle.

[2 marks]

[2 marks]

[3 marks]

Markschemecorrect approach (A1)

eg ,

5.3479

distance = 5.35 (m) A2 N3

[3 marks]

∫ 50 |v (t)|dt ∫2.95

0 (−v (t)) dt + ∫5

295 v (t) dt

2a.

Let , be a periodic function with

The following diagram shows the graph of .

There is a maximum point at A. The minimum value of is −13 .

Find the coordinates of A.

Markscheme−0.394791,13

A(−0.395, 13) A1A1 N2

[2 marks]

f (x) = 12 cos x − 5 sin x, −π ⩽ x ⩽ 2π f (x) = f (x + 2π)f

f

2b. For the graph of , write down the amplitude.f

[2 marks]

[1 mark]

Markscheme13 A1 N1

[1 mark]

2c. For the graph of , write down the period.

Markscheme, 6.28 A1 N1

[1 mark]

f

2π

2d. Hence, write in the form .


eg recognizing that amplitude is p or shift is r

(accept p = 13, r = 0.395) A1A1 N3

Note: Accept any value of r of the form

[3 marks]

f (x) p cos (x + r)

f (x) = 13 cos (x + 0.395)

0.395 + 2πk, k ∈ Z

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so

[1 mark]

[3 marks]

2e.

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, sothat it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by

Find the maximum speed of the ball.

Markschemerecognizing need for d ′(t) (M1)eg −12 sin(t) − 5 cos( t)

correct approach (accept any variable for t) (A1)

eg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32maximum speed = 13 (cms ) A1 N2

[3 marks]

d (t) = f (t) + 17, 0 ⩽ t ⩽ 5.

−1

2f. Find the first time when the ball’s speed is changing at a rate of 2 cm s .

Markschemerecognizing that acceleration is needed (M1)

eg a(t), d "(t)correct equation (accept any variable for t) (A1)

eg

valid attempt to solve their equation (M1)

eg sketch, 1.33

1.02154

1.02 A2 N3

[5 marks]

−2

a (t) = −2, ∣∣ (d′ (t))∣∣ = 2, −12 cos (t) + 5 sin (t) = −2ddt

[3 marks]

[5 marks]

3a.

A particle P moves along a straight line. The velocity v m s of P after t seconds is given by v (t)= 7 cos t − 5t , for 0 ≤ t ≤ 7.The following diagram shows the graph of v.

Find the initial velocity of P.

Markschemeinitial velocity when t = 0 (M1)

eg v(0)

v = 17 (m s ) A1 N2

[2 marks]

−1

cos t

−1

3b. Find the maximum speed of P.

Markschemerecognizing maximum speed when is greatest (M1)

eg minimum, maximum, v' = 0

one correct coordinate for minimum (A1)

eg 6.37896, −24.6571

24.7 (ms ) A1 N2

[3 marks]

|v|

−1

3c. Write down the number of times that the acceleration of P is 0 m s .−2

[2 marks]

[3 marks]

[3 marks]

Markschemerecognizing a = v ′ (M1)eg , correct derivative of first term

identifying when a = 0 (M1)

eg turning points of v, t-intercepts of v ′3 A1 N3

[3 marks]

a = dvdt

3d. Find the acceleration of P when it changes direction.

Markschemerecognizing P changes direction when v = 0 (M1)

t = 0.863851 (A1)

−9.24689

a = −9.25 (ms ) A2 N3

[4 marks]

−2

3e. Find the total distance travelled by P.

Markschemecorrect substitution of limits or function into formula (A1)eg

63.8874

63.9 (metres) A2 N3

[3 marks]

∫ 70 |v | , ∫0.8638

0 vdt − ∫7

0.8638 vdt, ∫ |7 cos x − 5xcos x |dx, 3.32 = 60.6

4a.

Note: In this question, distance is in metres and time is in seconds.

A particle P moves in a straight line for five seconds. Its acceleration at time is given by , for .

Write down the values of when .

ta = 3t2 − 14t + 8 0 ⩽ t ⩽ 5

t a = 0

[4 marks]

[3 marks]

[2 marks]

Markscheme A1A1 N2

[2 marks]

t = (exact), 0.667, t = 423

4b. Hence or otherwise, find all possible values of for which the velocity of P isdecreasing.

Markschemerecognizing that is decreasing when is negative (M1)

eg , sketch of

correct interval A1 N2

eg

[2 marks]

t

v a

a < 0, 3t2 − 14t + 8 ⩽ 0 a

< t < 423

4c.

When , the velocity of P is .

Find an expression for the velocity of P at time .

Markschemevalid approach (do not accept a definite integral) (M1)

eg

correct integration (accept missing ) (A1)(A1)(A1)

substituting , (must have ) (M1)

eg

A1 N6

[6 marks]

t = 0 3 m s−1

t

v ∫ a

c

t3 − 7t2 + 8t + c

t = 0, v = 3 c

3 = 03 − 7(02) + 8(0) + c, c = 3

v = t3 − 7t2 + 8t + 3

4d. Find the total distance travelled by P when its velocity is increasing.

[2 marks]

[6 marks]

[4 marks]

Markschemerecognizing that increases outside the interval found in part (b) (M1)

eg , diagram

one correct substitution into distance formula (A1)

eg

one correct pair (A1)

eg 3.13580 and 11.0833, 20.9906 and 35.2097

14.2191 A1 N2

[4 marks]

v

0 < t < , 4 < t < 523

∫0 |v| , ∫5

4 |v|, ∫4 |v|, ∫ 50 |v|

23

23

d = 14.2 (m)

5a.

A particle P moves along a straight line. Its velocity after seconds is given by , for . The following diagram shows the graph of .

Write down the first value of at which P changes direction.

Markscheme A1 N1

[1 mark]

vP m s−1 tvP = √t sin( t)π2 0 ⩽ t ⩽ 8 vP

t

t = 2

5b. Find the total distance travelled by P, for .0 ⩽ t ⩽ 8

[1 mark]

[2 marks]

Markschemesubstitution of limits or function into formula or correct sum (A1)

eg

9.64782

distance A1 N2

[2 marks]

∫ 80 |v|dt, ∫ ∣∣vQ∣∣dt, ∫2

0 vdt − ∫4

2 vdt + ∫6

4 vdt − ∫8

6 vdt

= 9.65 (metres)

5c. A second particle Q also moves along a straight line. Its velocity, after seconds is given by for . After seconds Q has travelled the same totaldistance as P.

Find .


eg

correct integration (A1)

eg

equating their expression to the distance travelled by their P (M1)

eg

5.93855

5.94 (seconds) A1 N3

[4 marks]

vQ m s−1 tvQ = √t 0 ⩽ t ⩽ 8 k

k

s = ∫ √t, ∫ k0 √tdt, ∫k

0∣∣vQ∣∣dt

∫ √t = t + c, [ x ]k0, k23

32 2

3

32 2

3

32

k = 9.65, ∫ k0 √tdt = 9.6523

32

Note: In this question, distance is in metres and time is in seconds.

[4 marks]

6. Note: In this question, distance is in metres and time is in seconds.

A particle moves along a horizontal line starting at a fixed point A. The velocity of the particle,at time , is given by , for . The following diagram shows the graph of

There are -intercepts at and .

Find the maximum distance of the particle from A during the time and justify youranswer.

v

t v(t) = 2t2−4t

t2−2t+2 0 ⩽ t ⩽ 5 v

t (0, 0) (2, 0)

0 ⩽ t ⩽ 5

[6 marks]

MarkschemeMETHOD 1 (displacement)

recognizing (M1)

consideration of displacement at and (seen anywhere) M1

eg and

Note: Must have both for any further marks.

correct displacement at and (seen anywhere) A1A1

(accept 2.28318), 1.55513

valid reasoning comparing correct displacements R1

eg , more left than right

2.28 (m) A1 N1

Note: Do not award the final A1 without the R1.

METHOD 2 (distance travelled)

recognizing distance (M1)

consideration of distance travelled from to 2 and to 5 (seen anywhere) M1

eg and

Note: Must have both for any further marks

correct distances travelled (seen anywhere) A1A1

2.28318, (accept ), 3.83832

valid reasoning comparing correct distance values R1

eg

2.28 (m) A1 N1

Note: Do not award the final A1 without the R1.

[6 marks]

s = ∫ vdt

t = 2 t = 5

∫ 20 v ∫5

0 v

t = 2 t = 5

−2.28318

|−2.28| > |1.56|

= ∫ |v|dt

t = 0 t = 2

∫ 20 v ∫5

2 v

−2.28318

3.84 − 2.28 < 2.28, 3.84 < 2 × 2.28

cm s−1

7a.

A particle P starts from a point A and moves along a horizontal straight line. Its velocity after seconds is given by


Find the initial velocity of .

Markschemevalid attempt to substitute into the correct function (M1)

eg

2 A1 N2

[2 marks]

v cm s−1t

v(t) = { −2t + 2, for 0 ⩽ t ⩽ 13√t + − 7, for 1 ⩽ t ⩽ 124

t2

v

P

t = 0

−2(0) + 2

7b.

P is at rest when and .

Find the value of .

Markschemerecognizing when P is at rest (M1)

5.21834

A1 N2

[2 marks]

t = 1 t = p

p

v = 0

p = 5.22 (seconds)

7c.

When , the acceleration of P is zero.

(i) Find the value of .

(ii) Hence, find the speed of P when .

t = q

q

t = q

[2 marks]

[2 marks]

[4 marks]

Markscheme(i) recognizing that (M1)

eg , minimum on graph

1.95343

A1 N2

(ii) valid approach to find their minimum (M1)

eg , reference to min on graph

1.75879

speed A1 N2

[4 marks]

a = v′

v′ = 0

q = 1.95

v(q), − 1.75879

= 1.76 (c m s−1)

7d. (i) Find the total distance travelled by P between and .

(ii) Hence or otherwise, find the displacement of P from A when .

Markscheme(i) substitution of correct into distance formula, (A1)

eg

4.45368

distance A1 N2

(ii) displacement from to (seen anywhere) (A1)

eg

displacement from to (A1)

eg

valid approach to find displacement for M1

eg

displacement A1 N2

[6 marks]

t = 1 t = p

t = p

v(t)

∫ p1 ∣∣3√t + − 7∣∣dt, ∣∣∫ 3√t + − 7dt∣∣4t2

4t2

= 4.45 (cm)

t = 1 t = p

−4.45368, ∫ p1 (3√t + − 7) dt4t2t = 0 t = 1

∫ 10 (−2t + 2)dt, 0.5 × 1 × 2, 1

0 ⩽ t ⩽ p

∫ 10 (−2t + 2)dt + ∫p

1 (3√t + − 7) dt, ∫ 10 (−2t + 2)dt − 4.454t2−3.45368

= −3.45 (cm)

8a.

A particle P moves along a straight line so that its velocity, , after seconds, is given by , for . The initial displacement of P from a fixed point O is 4

metres.

Find the displacement of P from O after 5 seconds.

v ms−1 tv = cos 3t − 2 sin t − 0.5 0 ⩽ t ⩽ 5

[6 marks]

[5 marks]

MarkschemeMETHOD 1

recognizing (M1)

recognizing displacement of P in first 5 seconds (seen anywhere) A1

(accept missing )

eg

valid approach to find total displacement (M1)

eg

0.284086

0.284 (m) A2 N3

METHOD 2

recognizing (M1)

correct integration A1

eg (do not penalize missing “ ”)

attempt to find (M1)

eg

attempt to substitute into their expression with (M1)

eg

0.284086

0.284 (m) A1 N3

[5 marks]

s = ∫ v

dt

∫ 50 vdt, − 3.71591

4 + (−3.7159), s = 4 + ∫ 50 v

s = ∫ v

sin 3t + 2 cos t − + c13t2 c

c

4 = sin(0) + 2 cos(0) − − + c, 4 = sin 3t + 2 cos t − + c, 2 + c = 41302

13

t2

t = 5 c

s(5), sin(15) + 2 cos(5)5 − − + 21352

8b.

The following sketch shows the graph of .

Find when P is first at rest.

v

[2 marks]

Markschemerecognizing that at rest, (M1)

A1 N2

[2 marks]

v = 0

t = 0.179900

t = 0.180 (secs)

8c. Write down the number of times P changes direction.

Markschemerecognizing when change of direction occurs (M1)

eg crosses axis

2 (times) A1 N2

[2 marks]

v t

8d. Find the acceleration of P after 3 seconds.

Markschemeacceleration is (seen anywhere) (M1)

eg

0.743631

A1 N2

[2 marks]

v′

v′(3)

0.744 (ms−2)

8e. Find the maximum speed of P.

Markschemevalid approach involving max or min of (M1)

eg , graph

one correct co-ordinate for min (A1)

eg

A1 N2

[3 marks]

v

v′ = 0, a = 0

1.14102, − 3.27876

3.28 (ms−1)

m s−1

[2 marks]

[2 marks]

[3 marks]

9. A particle moves in a straight line. Its velocity after seconds is given by

After seconds, the particle is 2 m from its initial position. Find the possible values of .


eg

correct integration (A1)

eg

recognizing that there are two possibilities (M1)

eg 2 correct answers,

two correct equations in A1A1

eg

0.42265, 1.57735

A1A1 N3

[7 marks]

v m s−1 t

v = 6t − 6, for 0 ⩽ t ⩽ 2.

p p

s = ∫ v, ∫ p0 6t − 6dt

∫ 6t − 6dt = 3t2 − 6t + C, [3t2 − 6t]p0

s = ±2, c ± 2

p

3p2 − 6p = 2, 3p2 − 6p = −2

p = 0.423 or p = 1.58

10a.

The velocity of a particle after seconds is given by

, for


Find the value of when the particle is at rest.

Markschemerecognizing particle at rest when (M1)

eg , -intercept on graph of

A2 N3

[3 marks]

v ms−1 t

v(t) = (0.3t + 0.1)t − 4 0 ≤ t ≤ 5

v

t

v = 0

(0.3t + 0.1)t − 4 = 0 x v

t = 4.27631

t = 4.28 (seconds)

0

[7 marks]

[3 marks]

10b. Find the value of when the acceleration of the particle is .

Markschemevalid approach to find when is (M1)

eg , minimum

A2 N3

[3 marks]

Total [6 marks]

t 0

t a 0

v′(t) = 0 v

t = 1.19236

t = 1.19 (seconds)

11.

A particle starts from point and moves along a straight line. Its velocity, , after seconds is given by , for . The particle is at rest when .


Find the distance travelled by the particle for .

Markschemecorrect substitution of function and/or limits into formula (A1)

(accept absence of d , but do not accept any errors)

eg

distance is A1 N2

[2 marks]

A v ms−1 tv(t) = e cost − 1

12 0 ≤ t ≤ 4 t = π2

v

0 ≤ t ≤ π2

t

∫0 v, ∫∣∣e

cost − 1∣∣dt, ∫ (e cost − 1)π

212

12

0.613747

0.614 [0.613, 0.614] (m)

Ramiro and Lautaro are travelling from Buenos Aires to El Moro.

[3 marks]

[2 marks]

12. Ramiro and Lautaro are travelling from Buenos Aires to El Moro.

Ramiro travels in a vehicle whose velocity in is given by , where is inseconds.

Lautaro travels in a vehicle whose displacement from Buenos Aires in metres is given by .

When , both vehicles are at the same point.

Find Ramiro’s displacement from Buenos Aires when .

MarkschemeMETHOD 1

(seen anywhere) (A1)

recognizing need to integrate (M1)

eg

correct expression A1A1

eg

Note: Award A1 for , and A1 for .

equate displacements to find C (R1)

eg

A1

attempt to find displacement (M1)

eg

A1 N5

METHOD 2


eg

valid approach involving a definite integral (M1)

eg

correct expression with limits (A1)

eg

A2



eg

ms−1 VR = 40 − t2 t

SL = 2t2 + 60

t = 0

t = 10

SL(0) = 60

VR

SR(t) ∫ VRdt

40t − t3 + C13

40t − t313

40(0) − (0)3 + C = 60, SL(0) = SR(0)13C = 60

SR(10), 40(10) − (10)3 + 6013126.666

126 (exact), 127 (m)23

VR

SR(t) = ∫ VRdt

∫ ba

VRdt

∫ 100 (40 − t2) dt, ∫ 100 VRdt, [40t − t

3]100

13

66.6666

SL(0) = 60

60 + 66.666

126.666

2

[8 marks]

(exact), (m) A1 N5

METHOD 3



eg


eg

Note: Award A1 for , and A1 for .

correct expression for Ramiro displacement A1

eg

A1


eg

(exact), 127 (m) A1 N5

[8 marks]

126 23 127

SL(0) = 60

VR

SR(t) = ∫ VRdt

40t − t3 + C13

40t − t313

SR(10) − SR(0), [40t − t3 + C]100

13

66.6666

60 + 66.6666

126 23

13a.

A particle moves in a straight line. Its velocity,

, at time

seconds, is given by

Find the velocity of the particle when .

Markschemesubstituting into (M1)

eg

velocity A1 N2

[2 marks]

v ms−1

t

v = (t2 − 4)3, for 0 ⩽ t ⩽ 3.

t = 1

t = 1 v

v(1), (12 − 4)3

= −27 (ms−1)

13b. Find the value of for which the particle is at rest.t

[2 marks]

[3 marks]

Markschemevalid reasoning (R1)

eg

correct working (A1)

eg , sketch

A1 N2

[3 marks]

v = 0, (t2 − 4)3 = 0

t2 − 4 = 0, t = ±2

t = 2

13c. Find the total distance the particle travels during the first three seconds.

Markschemecorrect integral expression for distance (A1)

eg ,

(do not accept )

A2 N3

[3 marks]

∫ 30 |v| , ∫∣∣(t

2 − 4)3∣∣ , − ∫2

0 vdt + ∫3

2 vdt

∫ 20 (4 − t2)3dt + ∫ 32 (t

2 − 4)3dt ∫ 30 vdt

86.2571

distance = 86.3 (m)

13d. Show that the acceleration of the particle is given by .

Markschemeevidence of differentiating velocity (M1)

eg

A2

AG N0

[3 marks]

a = 6t(t2 − 4)2

v′(t)

a = 3(t2 − 4)2(2t)

a = 6t(t2 − 4)2

13e. Find all possible values of for which the velocity and acceleration are both positiveor both negative.

t

[3 marks]

[3 marks]

[4 marks]

MarkschemeMETHOD 1

valid approach M1

eg graphs of and


eg areas of same sign indicated on graph

(accept ) A2 N2

METHOD 2

recognizing that (accept is always positive) (seen anywhere) R1

recognizing that is positive when (seen anywhere) (R1)

(accept ) A2 N2

[4 marks]

v a

2 < t ⩽ 3 t > 2

a ⩾ 0 av t > 2

2 < t ⩽ 3 t > 2

14a.

A particle moves along a straight line such that its velocity,

, is given by

, for

.

On the grid below, sketch the graph of , for .

v ms−1

v(t) = 10te−1.7t

t ⩾ 0

v 0 ⩽ t ⩽ 4 [3 marks]

Markscheme

A1A2 N3

Notes: Award A1 for approximately correct domain .

The shape must be approximately correct, with maximum skewed left. Only if the shape isapproximately correct, award A2 for all the following approximately correct features, incircle of tolerance where drawn (accept seeing correct coordinates for the maximum, evenif point outside circle):

Maximum point, passes through origin, asymptotic to -axis (but must not touch the axis).

If only two of these features are correct, award A1.

[3 marks]

0 ⩽ t ⩽ 4

t

14b. Find the distance travelled by the particle in the first three seconds.

Markschemevalid approach (including and ) (M1)

eg , area from to (may be shaded in diagram)

A1 N2

[2 marks]

0 3

∫ 30 10te−1.7tdt, ∫ 30 f(x) 0 3

distance = 3.33 (m)

14c. Find the velocity of the particle when its acceleration is zero.

[2 marks]

[3 marks]

Markschemerecognizing acceleration is derivative of velocity (R1)

eg , attempt to find , reference to maximum on the graph of

valid approach to find when (may be seen on graph) (M1)

eg

A1 N3

Note: Award R1M1A0 for if velocity is not identified as final answer

[3 marks]

a = dvdtdvdt v

v a = 0

= 0, 10e−1.7t − 17te−1.7t = 0, t = 0.588dvdtvelocity = 2.16 (ms−1)

(0.588,216)

15. A rocket moving in a straight line has velocity km s and displacement km at time seconds. The velocity is given by . When , .

Find an expression for the displacement of the rocket in terms of .

Markschemeevidence of anti-differentiation (M1)

eg

A2A1

Note: Award A2 for , A1 for .

attempt to substitute ( , ) into their integrated expression (even if is missing) (M1)


eg ,

A1 N6

Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration whichmust involve a power of .

[7 marks]

v –1 st v v(t) = 6e2t + t t = 0 s = 10

t

∫ (6e2t + t)

s = 3e2t + + Ct2

2

3e2t t2

2

0 10 C

10 = 3 + C C = 7

s = 3e2t + + 7t2

2

e

The velocity of a particle in ms is given by−1

[7 marks]

16a.

The velocity of a particle in ms is given by

, for

.

On the grid below, sketch the graph of .

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape crossing x-axis with .

Only if this A1 is awarded, award the following:

A1 for maximum in circle, A1 for endpoints in circle.

[3 marks]

−1

v = esin t − 10 ≤ t ≤ 5

v

3 < x < 3.5

16b. Find the total distance travelled by the particle in the first five seconds.

Markscheme (exact), A1 N1

[1 mark]

t = π 3.14

16c. Write down the positive -intercept.t

[3 marks]

[1 mark]

[4 marks]

Markschemerecognizing distance is area under velocity curve (M1)

eg , shading on diagram, attempt to integrate

valid approach to find the total area (M1)

eg , , ,

correct working with integration and limits (accept or missing ) (A1)

eg , ,

distance (m) A1 N3

[4 marks]

s = ∫ v

area A + area B ∫ vdt − ∫ vdt ∫ 3.140 vdt+ ∫5

3.14 vdt ∫ |v|

dx dt

∫ 3.140 vdt+ ∫3.14

5 vdt 3.067… + 0.878… ∫5

0∣∣esin t − 1∣∣

= 3.95

17a.

A particle’s displacement, in metres, is given by

, for

, where t is the time in seconds.

On the grid below, sketch the graph of .

s(t) = 2t cos t0 ≤ t ≤ 6

s [4 marks]

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct shape (do not accept line segments).

Only if this A1 is awarded, award the following:

A1 for maximum and minimum within circles,

A1 for x-intercepts between 1 and 2 and between 4 and 5,

A1 for left endpoint at and right endpoint within circle.

[4 marks]

(0, 0)

17b. Find the maximum velocity of the particle.

Markschemeappropriate approach (M1)

e.g. recognizing that , finding derivative,

valid method to find maximum (M1)

e.g. sketch of , ,

A1 N2

[3 marks]

v = s′ a = s′′

v v′(t) = 0 t = 5.08698…

v = 10.20025 …

v = 10.2 [10.2, 10.3]

In this question, you are given that

[3 marks]

18a.

In this question, you are given that

, and

.

The displacement of an object from a fixed point, O is given by

for

.

Find .

Markscheme A1A2 N3

Note: Award A1 for 1, A2 for .

[3 marks]

cos =π312

sin =π3√32

s(t) = t − sin 2t0 ≤ t ≤ π

s′(t)

s′(t) = 1 − 2 cos 2t

−2 cos 2t

18b. In this interval, there are only two values of t for which the object is not moving. Onevalue is .

Find the other value.

Markschemeevidence of valid approach (M1)

e.g. setting

correct working A1

e.g. ,

, , (A1)

A1 N3

[4 marks]

t = π6

s′(t) = 0

2 cos 2t = 1 cos 2t = 12

2t = π35π3 …

t = 5π6

18c. Show that between these two values of t .s′(t) > 0

[3 marks]

[4 marks]

[3 marks]

Markschemeevidence of valid approach (M1)

e.g. choosing a value in the interval

correct substitution A1

e.g.

A1

AG N0

[3 marks]

< t 0

18d. Find the distance travelled between these two values of t .

Markschemeevidence of approach using s or integral of (M1)

e.g. ; , ;

substituting values and subtracting (M1)

e.g. ,


e.g. ,

distance is A1A1 N3

Note: Award A1 for , A1 for .

[5 marks]

s′

∫ s′(t)dt s ( )5π6 s ( )π6 [t − sin 2t]

5π6

π

6

s ( ) − s ( )5π6π6 ( − ) − ( − (− ))π6 √32 5π6 √32

− sin − [ − sin ]5π65π3

π6

π3 ( − (− )) − ( − )5π6 √32 π6 √32

+ √32π32π3 √3

19a.

A particle moves in a straight line with velocity

, for

, where v is in centimetres per second and t is in seconds.

Find the acceleration of the particle after 2.7 seconds.

v = 12t − 2t3 − 1t ≥ 0

[5 marks]

[3 marks]

Markschemerecognizing that acceleration is the derivative of velocity (seen anywhere) (R1)

e.g.

correctly substituting 2.7 into their expression for a (not into v) (A1)

e.g.

(exact), A1 N3

[3 marks]

a = ,v′,12 − 6t2d2s

dt2

s′′(2.7)

acceleration = −31.74 −31.7

19b. Find the displacement of the particle after 1.3 seconds.

Markschemerecognizing that displacement is the integral of velocity R1

e.g.

correctly substituting 1.3 (A1)

e.g.

(exact), (cm) A1 N2

[3 marks]

s = ∫ v

∫ 1.30 vdt

displacement = 7.41195 7.41

20a.

Let

, where

. The function v is obtained when the graph of f is transformed by

a stretch by a scale factor of

parallel to the y-axis,

followed by a translation by the vector

.

Find , giving your answer in the form .

f(t) = 2t2 + 7t > 0

13

( 2−4

)

v(t) a(t − b)2 + c

[3 marks]

[4 marks]

Markschemeapplies vertical stretch parallel to the y-axis factor of (M1)

e.g. multiply by , ,

applies horizontal shift 2 units to the right (M1)

e.g. ,

applies a vertical shift 4 units down (M1)

e.g. subtracting 4, ,

A1 N4

[4 marks]

13

13 f(t)

13 × 2

13

f(t − 2) t − 2

f(t) − 4 − 473

v(t) = (t − 2)2 −2353

20b. A particle moves along a straight line so that its velocity in ms , at time t seconds, isgiven by v . Find the distance the particle travels between and .

Markschemerecognizing that distance travelled is area under the curve M1

e.g. , sketch

distance = 15.576 (accept 15.6) A2 N2

[3 marks]

−1

t = 5.0 t = 6.8

∫ v, (t − 2)3 − t2953

21a.

The velocity v ms of a particle at time t seconds, is given by

, for

.

Write down the velocity of the particle when .

Markscheme A1 N1

[1 mark]

−1

v = 2t + cos 2t0 ≤ t ≤ 2

t = 0

v = 1

[3 marks]

[1 mark]

21b. When , the acceleration is zero.

(i) Show that .

(ii) Find the exact velocity when .

Markscheme(i) A1

A1A1

Note: Award A1 for coefficient 2 and A1 for .

evidence of considering acceleration = 0 (M1)

e.g. ,

correct manipulation A1

e.g. ,

(accept ) A1

AG N0

(ii) attempt to substitute into v (M1)

e.g.

A1 N2

[8 marks]

t = k

k = π4t = π4

(2t) = 2ddt

(cos 2t) = −2 sin 2tddt

− sin 2t

= 0dvdt 2 − 2 sin 2t = 0

sin 2k = 1 sin 2t = 1

2k = π2 2t =π2

k = π4t = π4

2 ( ) + cos( )π42π4

v = π2

21c. When , and when , .

Sketch a graph of v against t .

Markscheme

A1A1A2 N4

Notes: Award A1 for y-intercept at , A1 for curve having zero gradient at , A2for shape that is concave down to the left of and concave up to the right of . If a correctcurve is drawn without indicating , do not award the second A1 for the zero gradient,but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essentialfeatures need to be clear.

[4 marks]

t < π4 > 0dvdt t >

π4 > 0

dvdt

(0, 1) t = π4π4

π4

t = π4

[8 marks]

[4 marks]

21d. Let d be the distance travelled by the particle for .

(i) Write down an expression for d .

(ii) Represent d on your sketch.

Markscheme(i) correct expression A2

e.g. , , ,

(ii)

A1

Note: The line at needs to be clearly after .

[3 marks]

0 ≤ t ≤ 1

∫ 10 (2t + cos 2t)dt [t2 + ]

10

sin 2t2 1 +

sin 22 ∫

10 vdt

t = 1 t = π4

22a.

The following diagram shows part of the graph of a quadratic function f .

The x-intercepts are at

and

, and the y-intercept is at

.

Write down in the form .

(−4, 0)(6, 0)(0, 240)

f(x) f(x) = −10(x − p)(x − q)

[3 marks]

[2 marks]

Markscheme A1A1 N2

[2 marks]

f(x) = −10(x + 4)(x − 6)

22b. Find another expression for in the form .

MarkschemeMETHOD 1

attempting to find the x-coordinate of maximum point (M1)

e.g. averaging the x-intercepts, sketch, , axis of symmetry

attempting to find the y-coordinate of maximum point (M1)

e.g.

A1A1 N4

METHOD 2

attempt to expand (M1)

e.g.

attempt to complete the square (M1)

e.g.

A1A1 N4

[4 marks]

f(x) f(x) = −10(x − h)2 + k

y′ = 0

k = −10(1 + 4)(1 − 6)

f(x) = −10(x − 1)2 + 250

f(x)

−10(x2 − 2x − 24)

−10((x − 1)2 − 1 − 24)

f(x) = −10(x − 1)2 + 250

22c. Show that can also be written in the form .

Markschemeattempt to simplify (M1)

e.g. distributive property,

correct simplification A1

e.g. ,

AG N0

[2 marks]

f(x) f(x) = 240 + 20x − 10x2

−10(x − 1)(x − 1) + 250

−10(x2 − 6x + 4x − 24) −10(x2 − 2x + 1) + 250

f(x) = 240 + 20x − 10x2

ms−1

[4 marks]

[2 marks]

22d. A particle moves along a straight line so that its velocity, , at time t seconds isgiven by , for .

(i) Find the value of t when the speed of the particle is greatest.

(ii) Find the acceleration of the particle when its speed is zero.

Markscheme(i) valid approach (M1)

e.g. vertex of parabola,

A1 N2

(ii) recognizing (M1)

A1A1

speed is zero (A1)

( ) A1 N3

[7 marks]

v ms−1v = 240 + 20t − 10t2 0 ≤ t ≤ 6

v′(t) = 0

t = 1

a(t) = v′(t)

a(t) = 20 − 20t

⇒ t = 6

a(6) = −100 ms−2

23a.

The velocity v ms of an object after t seconds is given by

, for

.

On the grid below, sketch the graph of v , clearly indicating the maximum point.

−1

v(t) = 15√t − 3t0 ≤ t ≤ 25

[7 marks]

[3 marks]

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape, A1 for right endpoint at and A1for maximum point in circle.

[3 marks]

(25, 0)

23b. (i) Write down an expression for d .

(ii) Hence, write down the value of d .

Markscheme(i) recognizing that d is the area under the curve (M1)

e.g.

correct expression in terms of t, with correct limits A2 N3

e.g. ,

(ii) (m) (accept 149 to 3 sf) A1 N1

[4 marks]

∫ v(t)

d = ∫ 90 (15√t − 3t)dt d = ∫9

0 vdt

d = 148.5

24. The acceleration, , of a particle at time t seconds is given by

The particle is at rest when .

Find the velocity of the particle when .

a ms−2

a = + 3 sin 2t, for t ≥ 1.1t

t = 1

t = 5

[4 marks]

[7 marks]

Markschemeevidence of integrating the acceleration function (M1)

e.g.


evidence of substituting (1, 0) (M1)

e.g.

(A1)

(A1)

(accept the exact answer ) A1 N3

[7 marks]

∫ ( + 3 sin 2t)dt1t

ln t − cos 2t + c32

0 = ln 1 − cos 2 + c32

c = −0.624 (= cos 2 − ln 1 or cos 2)3232

v = ln t − cos 2t − 0.62432(= ln t − cos 2t + cos 2 or lnt − cos 2t + cos 2 − ln 1)32

32

32

32

v(5) = 2.24 ln 5 − 1.5cos 10 + 1.5cos 2

25a.

The following diagram shows the graphs of the displacement, velocity and acceleration of a moving object as

functions of time, t.

Complete the following table by noting which graph A, B or C corresponds to eachfunction.

[4 marks]

Markscheme

A2A2 N4

[4 marks]

25b. Write down the value of t when the velocity is greatest.

Markscheme A2 N2

[2 marks]

t = 3

26a.

In this question s represents displacement in metres and t represents time in seconds.

The velocity v m s of a moving body is given by

where a is a non-zero constant.

(i) If when , find an expression for s in terms of a and t.

(ii) If when , write down an expression for s in terms of a and t.

MarkschemeNote: In this question, do not penalize absence of units.

(i) (M1)

(A1)(A1)

substituting when ( ) (M1)

A1 N5

(ii) A1 N1

[6 marks]

–1

v = 40 − at

s = 100 t = 0

s = 0 t = 0

s = ∫ (40 − at)dt

s = 40t − at2 + c12s = 100 t = 0 c = 100

s = 40t − at2 + 10012

s = 40t − at212

26b.

Trains approaching a station start to slow down when they pass a point P. As a train slows down, its velocity is given by

, where

at P. The station is 500 m from P.

A train M slows down so that it comes to a stop at the station.

(i) Find the time it takes train M to come to a stop, giving your answer in terms of a.

(ii) Hence show that .

v = 40 − att = 0

a = 85

[2 marks]

[6 marks]

[6 marks]

Markscheme(i) stops at station, so (M1)

(seconds) A1 N2

(ii) evidence of choosing formula for s from (a) (ii) (M1)

substituting (M1)

e.g.

setting up equation M1

e.g. , ,

evidence of simplification to an expression which obviously leads to A1

e.g. , ,

AG N0

[6 marks]

v = 0

t = 40a

t = 40a

40 × − a ×40a

12

402

a2

500 = s 500 = 40 × − a ×40a

12

402

a2500 = −1600

a800a

a = 85

500a = 800 5 = 8a

1000a = 3200 − 1600

a = 85

26c. For a different train N, the value of a is 4.

Show that this train will stop before it reaches the station.

[5 marks]

MarkschemeMETHOD 1

, stops when

(A1)

A1

substituting into expression for s M1

A1

since (allow FT on their s, if ) R1

train stops before the station AG N0

METHOD 2

from (b) A2

substituting into expression for s

e.g. M1

A1

since R1

train stops before the station AG N0

METHOD 3

a is deceleration A2

A1

so stops in shorter time (A1)

so less distance travelled R1

so stops before station AG N0

[5 marks]

v = 40 − 4t v = 0

40 − 4t = 0

t = 10

s = 40 × 10 − × 4 × 10212s = 200

200 < 500 s < 500

t = = 10404

s = 40 × 10 − × 4 × 10212s = 200

200 < 500

4 > 85

27a.

The acceleration,

, of a particle at time t seconds is given by

.

Find the acceleration of the particle at .

Markschemesubstituting (M1)

e.g.

A1 N2

[2 marks]

a ms−2

a = 2t + cos t

t = 0

t = 0

a(0) = 0 + cos 0

a(0) = 1

[2 marks]

27b. Find the velocity, v, at time t, given that the initial velocity of the particle is .

Markschemeevidence of integrating the acceleration function (M1)

e.g.


Note: If " " is omitted, award no further marks.

evidence of substituting (2,0) into indefinite integral (M1)

e.g. ,

A1 N3

[5 marks]

2 ms−1

∫ (2t + cos t)dt

t2 + sin t + c

+c

2 = 0 + sin 0 + c c = 2

v(t) = t2 + sin t + 2

27c. Find , giving your answer in the form .

Markscheme A1A1A1

Note: Award A1 for each correct term.

evidence of using (M1)


e.g. ,

(accept , ) A1A1 N3

[7 marks]

∫ 30 vdt p − qcos 3

∫ (t2 + sin t + 2)dt = − cos t + 2tt3

3

v(3) − v(0)

(9 − cos 3 + 6) − (0 − cos 0 + 0) (15 − cos 3) − (−1)

16 − cos 3 p = 16 q = −1

27d. What information does the answer to part (c) give about the motion of the particle?

Markschemereference to motion, reference to first 3 seconds R1R1 N2

e.g. displacement in 3 seconds, distance travelled in 3 seconds

[2 marks]

28. A particle moves along a straight line so that its velocity, at time t seconds isgiven by . When , the displacement, s, of the particle is 7 metres. Find anexpression for s in terms of t.

v ms−1v = 6e3t + 4 t = 0

[5 marks]

[7 marks]

[2 marks]

[7 marks]

Markschemeevidence of anti-differentiation (M1)

e.g.

A2A1

substituting , (M1)

A1

A1 N3

[7 marks]

s = ∫ (6e3x + 4)dx

s = 2e3t + 4t + C

t = 0

7 = 2 + C

C = 5

s = 2e3t + 4t + 5

29a.

A toy car travels with velocity v ms for six seconds. This is shown in the graph below.

Write down the car’s velocity at .

Markscheme A1 N1

[1 mark]

−1

t = 3

4 (ms−1)

29b. Find the car’s acceleration at .

Markschemerecognizing that acceleration is the gradient M1

e.g.

A1 N1

[2 marks]

t = 1.5

a(1.5) = 4−02−0

a = 2 (ms−2)

[1 mark]

[2 marks]

Printed for GEMS INTERNATIONAL SCHOOL AL KHAIL

© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

29c. Find the total distance travelled.

Markschemerecognizing area under curve M1

e.g. trapezium, triangles, integration


e.g. ,

distance 18 (m) A1 N2

[3 marks]

(3 + 6)412 ∫6

0 |v(t)|dt

[3 marks]

Kinematics [285 marks]MarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkschemeMarkscheme

kinematics [285 marks] - github pages[2 marks] t=2 (exact), 0.667, t=4 3 4b.hence or otherwise, find...

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