IBDP Maths AA: Topic SL 5.9: Kinematic problems involving displacement: IB style Questions SL Paper 1

Question

In this question, all lengths are in metres and time is in seconds.

Consider two particles, P1 and P2 , which start to move at the same time.

Particle P1 moves in a straight line such that its displacement from a fixed-point is given by s (t) = 10 – $$\frac{7}{4}$$ t2 , for t ≥ 0 .

1. Find an expression for the velocity of P1 at time t . [2]

Particle P2 also moves in a straight line. The position of P2 is given by

The speed of P1 is greater than the speed of P2 when t > q .

2. Find the value of q . [5]

Answer/Explanation

Ans:

(a)

recognizing velocity is derivative of displacement

eg v= $$\frac{ds}{dt},\frac{d}{dt}(10-\frac{7}{4}t^{2})$$

velocity = -$$\frac{14}{4}t (=-\frac{7}{2}t)$$

(b)

valid approach to find speed of P2

$$|(_{-3}^{4})|, \sqrt{4^{2}+(-3)^{2}}$$, volocity = $$\sqrt{4^{2}+(-3)^{2}}$$

correct speed

eg $$5ms ^{-1}$$

recognizing relationship between speed and velocity (may be seen in inequality/equation)

correct inequality or equation that compares speed or velocity (accept any variable for q)

eg $$|-\frac{7}{2}t|>5,-\frac{7}{2}q<-5 , \frac{7}{2}q> 5,\frac{7}{2}q= 5$$

$$q= \frac{10}{7}$$(seconds)(accept t $$> \frac{10}{7})$$, do not accept $$t = \frac{10}{7}$$

Question

A particle moves along a straight line so that its velocity, $$v{\text{ m}}{{\text{s}}^{ – 1}}$$at time t seconds is given by $$v = 6{{\rm{e}}^{3t}} + 4$$ . When $$t = 0$$ , the displacement, s, of the particle is 7 metres. Find an expression for s in terms of t.

Answer/Explanation

Markscheme

evidence of anti-differentiation     (M1)

e.g. $$s = \int {(6{{\rm{e}}^{3x}} + 4)} {\rm{d}}x$$

$$s = 2{{\rm{e}}^{3t}} + 4t + C$$     A2A1

substituting $$t = 0$$ ,    (M1)

$$7 = 2 + C$$     A1

$$C = 5$$

$$s = 2{{\rm{e}}^{3t}} + 4t + 5$$    A1     N3

[7 marks]

Question

The acceleration, $$a{\text{ m}}{{\text{s}}^{ – 2}}$$, of a particle at time t seconds is given by $$a = 2t + \cos t$$ .

Find the acceleration of the particle at $$t = 0$$ .

[2]
a.

Find the velocity, v, at time t, given that the initial velocity of the particle is $${\text{m}}{{\text{s}}^{ – 1}}$$ .

[5]
b.

Find $$\int_0^3 {v{\rm{d}}t}$$ , giving your answer in the form $$p – q\cos 3$$ .

[7]
c.

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

[2]
d.
Answer/Explanation

Markscheme

substituting $$t = 0$$     (M1)

e.g. $$a(0) = 0 + \cos 0$$

$$a(0) = 1$$     A1     N2

[2 marks]

a.

evidence of integrating the acceleration function     (M1)

e.g. $$\int {(2t + \cos t){\text{d}}t}$$

correct expression $${t^2} + \sin t + c$$     A1A1

Note: If “$$+ c$$” is omitted, award no further marks.

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

e.g. $$2 = 0 + \sin 0 + c$$ , $$c = 2$$

$$v(t) = {t^2} + \sin t + 2$$     A1     N3

[5 marks]

b.

$$\int {({t^2} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} – \cos t + 2t$$     A1A1A1

Note: Award A1 for each correct term.

evidence of using $$v(3) – v(0)$$     (M1)

correct substitution     A1

e.g. $$(9 – \cos 3 + 6) – (0 – \cos 0 + 0)$$ , $$(15 – \cos 3) – ( – 1)$$

$$16 – \cos 3$$ (accept $$p = 16$$ , $$q = – 1$$ )     A1A1     N3

[7 marks]

c.

reference to motion, reference to first 3 seconds     R1R1     N2

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

[2 marks]

d.

Question

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 each function.

[4]
a.

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

[2]
b.
Answer/Explanation

Markscheme

A2A2     N4

[4 marks]

a.

$$t = 3$$     A2     N2

[2 marks]

b.

Question

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

The velocity v m s–1 of a moving body is given by $$v = 40 – at$$ where a is a non-zero constant.

Trains approaching a station start to slow down when they pass a point P. As a train slows down, its velocity is given by $$v = 40 – at$$ , where $$t = 0$$ at P. The station is 500 m from P.

(i)     If $$s = 100$$ when $$t = 0$$ , find an expression for s in terms of a and t.

(ii)    If $$s = 0$$ when $$t = 0$$ , write down an expression for s in terms of a and t.

[6]
a.

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 $$a = \frac{8}{5}$$ .

[6]
b.

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

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

[5]
c.
Answer/Explanation

Markscheme

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

(i) $$s = \int {(40 – at){\rm{d}}t}$$     (M1)

$$s = 40t – \frac{1}{2}a{t^2} + c$$     (A1)(A1)

substituting $$s = 100$$ when $$t = 0$$ ($$c = 100$$ )     (M1)

$$s = 40t – \frac{1}{2}a{t^2} + 100$$     A1     N5

(ii) $$s = 40t – \frac{1}{2}a{t^2}$$     A1     N1

[6 marks]

a.

(i) stops at station, so $$v = 0$$     (M1)

$$t = \frac{{40}}{a}$$ (seconds)     A1     N2

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

substituting $$t = \frac{{40}}{a}$$     (M1)

e.g. $$40 \times \frac{{40}}{a} – \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^2}}}$$

setting up equation     M1

e.g. $$500 = s$$ , $$500 = 40 \times \frac{{40}}{a} – \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^2}}}$$ , $$500 = \frac{{1600}}{a} – \frac{{800}}{a}$$

evidence of simplification to an expression which obviously leads to $$a = \frac{8}{5}$$     A1

e.g. $$500a = 800$$ , $$5 = \frac{8}{a}$$ , $$1000a = 3200 – 1600$$

$$a = \frac{8}{5}$$     AG     N0

[6 marks]

b.

METHOD 1

$$v = 40 – 4t$$ , stops when $$v = 0$$

$$40 – 4t = 0$$     (A1)

$$t = 10$$     A1

substituting into expression for s     M1

$$s = 40 \times 10 – \frac{1}{2} \times 4 \times {10^2}$$

$$s = 200$$     A1

since $$200 < 500$$ (allow FT on their s, if $$s < 500$$ )     R1

train stops before the station     AG     N0

METHOD 2

from (b) $$t = \frac{{40}}{4} = 10$$     A2

substituting into expression for s

e.g. $$s = 40 \times 10 – \frac{1}{2} \times 4 \times {10^2}$$     M1

$$s = 200$$     A1

since $$200 < 500$$      R1

train stops before the station     AG     N0

METHOD 3

a is deceleration     A2

$$4 > \frac{8}{5}$$    A1

so stops in shorter time     (A1)

so less distance travelled     R1

so stops before station     AG     N0

[5 marks]

c.

Question

The velocity v ms−1 of a particle at time t seconds, is given by $$v = 2t + \cos 2t$$ , for $$0 \le t \le 2$$ .

Write down the velocity of the particle when $$t = 0$$ .

[1]
a.

When $$t = k$$ , the acceleration is zero.

(i)     Show that $$k = \frac{\pi }{4}$$ .

(ii)    Find the exact velocity when $$t = \frac{\pi }{4}$$ .

[8]
b(i) and (ii).

When $$t < \frac{\pi }{4}$$ , $$\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0$$ and when $$t > \frac{\pi }{4}$$ , $$\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0$$  .

Sketch a graph of v against t .

[4]
c.

Let d be the distance travelled by the particle for $$0 \le t \le 1$$ .

(i)     Write down an expression for d .

(ii)    Represent d on your sketch.

[3]
d(i) and (ii).
Answer/Explanation

Markscheme

$$v = 1$$     A1     N1

[1 mark]

a.

(i) $$\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2$$     A1

$$\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t$$     A1A1

Note: Award A1 for coefficient 2 and A1 for $$– \sin 2t$$ .

evidence of considering acceleration = 0     (M1)

e.g. $$\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0$$ , $$2 – 2\sin 2t = 0$$

correct manipulation     A1

e.g. $$\sin 2k = 1$$ , $$\sin 2t = 1$$

$$2k = \frac{\pi }{2}$$ (accept $$2t = \frac{\pi }{2}$$ )     A1

$$k = \frac{\pi }{4}$$     AG     N0

(ii) attempt to substitute $$t = \frac{\pi }{4}$$ into v     (M1)

e.g. $$2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)$$

$$v = \frac{\pi }{2}$$     A1     N2

[8 marks]

b(i) and (ii).

A1A1A2      N4

Notes: Award A1 for y-intercept at $$(0{\text{, }}1)$$ , A1 for curve having zero gradient at $$t = \frac{\pi }{4}$$ , A2 for shape that is concave down to the left of $$\frac{\pi }{4}$$ and concave up to the right of $$\frac{\pi }{4}$$ . If a correct curve is drawn without indicating $$t = \frac{\pi }{4}$$ , 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 essential features need to be clear.

[4 marks]

c.

(i) correct expression     A2

e.g. $$\int_0^1 {(2t + \cos 2t){\rm{d}}t}$$ , $$\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1$$ , $$1 + \frac{{\sin 2}}{2}$$ , $$\int_0^1 {v{\rm{d}}t}$$

(ii)

A1

Note: The line at $$t = 1$$ needs to be clearly after $$t = \frac{\pi }{4}$$ .

[3 marks]

d(i) and (ii).

Question

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

The x-intercepts are at $$( – 4{\text{, }}0)$$ and $$(6{\text{, }}0)$$ , and the y-intercept is at $$(0{\text{, }}240)$$ .

Write down $$f(x)$$ in the form $$f(x) = – 10(x – p)(x – q)$$ .

[2]
a.

Find another expression for $$f(x)$$ in the form $$f(x) = – 10{(x – h)^2} + k$$ .

[4]
b.

Show that $$f(x)$$ can also be written in the form $$f(x) = 240 + 20x – 10{x^2}$$ .

[2]
c.

A particle moves along a straight line so that its velocity, $$v{\text{ m}}{{\text{s}}^{ – 1}}$$ , at time t seconds is given by $$v = 240 + 20t – 10{t^2}$$ , for $$0 \le t \le 6$$ .

(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.

[7]
d(i) and (ii).
Answer/Explanation

Markscheme

$$f(x) = – 10(x + 4)(x – 6)$$     A1A1     N2

[2 marks]

a.

METHOD 1

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

e.g. averaging the x-intercepts, sketch, $$y’ = 0$$ , axis of symmetry

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

e.g. $$k = – 10(1 + 4)(1 – 6)$$

$$f(x) = – 10{(x – 1)^2} + 250$$     A1A1     N4

METHOD 2

attempt to expand $$f(x)$$     (M1)

e.g. $$– 10({x^2} – 2x – 24)$$

attempt to complete the square     (M1)

e.g. $$– 10({(x – 1)^2} – 1 – 24)$$

$$f(x) = – 10{(x – 1)^2} + 250$$     A1A1     N4

[4 marks]

b.

attempt to simplify     (M1)

e.g. distributive property, $$– 10(x – 1)(x – 1) + 250$$

correct simplification     A1

e.g. $$– 10({x^2} – 6x + 4x – 24)$$ , $$– 10({x^2} – 2x + 1) + 250$$

$$f(x) = 240 + 20x – 10{x^2}$$     AG     N0

[2 marks]

c.

(i) valid approach     (M1)

e.g. vertex of parabola, $$v'(t) = 0$$

$$t = 1$$     A1     N2

(ii) recognizing $$a(t) = v'(t)$$     (M1)

$$a(t) = 20 – 20t$$     A1A1

speed is zero $$\Rightarrow t = 6$$     (A1)

$$a(6) = – 100$$ ($${\text{m}}{{\text{s}}^{ – 2}}$$)     A1     N3

[7 marks]

d(i) and (ii).

Question

In this question, you are given that $$\cos \frac{\pi }{3} = \frac{1}{2}$$ , and $$\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}$$ .

The displacement of an object from a fixed point, O is given by $$s(t) = t – \sin 2t$$ for $$0 \le t \le \pi$$ .

Find $$s'(t)$$ .

[3]
a.

In this interval, there are only two values of t for which the object is not moving. One value is $$t = \frac{\pi }{6}$$ .

Find the other value.

[4]
b.

Show that $$s'(t) > 0$$ between these two values of t .

[3]
c.

Find the distance travelled between these two values of t .

[5]
d.
Answer/Explanation

Markscheme

$$s'(t) = 1 – 2\cos 2t$$    A1A2     N3

Note: Award A1 for 1, A2 for $$- 2\cos 2t$$ .

[3 marks]

a.

evidence of valid approach     (M1)

e.g. setting $$s'(t) = 0$$

correct working     A1

e.g. $$2\cos 2t = 1$$ , $$\cos 2t = \frac{1}{2}$$

$$2t = \frac{\pi }{3}$$ , $$\frac{{5\pi }}{3}$$ , $$\ldots$$     (A1)

$$t = \frac{{5\pi }}{6}$$     A1     N3

[4 marks]

b.

evidence of valid approach     (M1)

e.g. choosing a value in the interval $$\frac{\pi }{6} < t < \frac{{5\pi }}{6}$$

correct substitution     A1

e.g. $$s’\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi$$

$$s’\left( {\frac{\pi }{2}} \right) = 3$$     A1

$$s'(t) > 0$$     AG     N0

[3 marks]

c.

evidence of approach using s or integral of $$s’$$     (M1)

e.g. $$\int {s'(t){\rm{d}}t}$$ ; $$s\left( {\frac{{5\pi }}{6}} \right)$$ , $$s\left( {\frac{\pi }{6}} \right)$$ ; $$\left[ {t – \sin 2t} \right]_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}}$$

substituting values and subtracting     (M1)

e.g. $$s\left( {\frac{{5\pi }}{6}} \right) – s\left( {\frac{\pi }{6}} \right)$$ , $$\left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right) – \left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right)$$

correct substitution     A1

e.g. $$\frac{{5\pi }}{6} – \sin \frac{{5\pi }}{3} – \left[ {\frac{\pi }{6} – \sin \frac{\pi }{3}} \right]$$ , $$\left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right) – \left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right)$$

distance is $$\frac{{2\pi }}{3} + \sqrt 3$$     A1A1     N3

Note: Award A1 for $$\frac{{2\pi }}{3}$$ , A1 for $$\sqrt 3$$ .

[5 marks]

d.

Question

A rocket moving in a straight line has velocity $$v$$ km s–1 and displacement $$s$$ km at time $$t$$ seconds. The velocity $$v$$ is given by $$v(t) = 6{{\rm{e}}^{2t}} + t$$ . When $$t = 0$$ , $$s = 10$$ .

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

Answer/Explanation

Markscheme

evidence of anti-differentiation     (M1)

eg   $$\int {(6{{\rm{e}}^{2t}} + t)}$$

$$s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + C$$     A2A1

Note: Award A2 for $$3{{\rm{e}}^{2t}}$$ , A1 for $$\frac{{{t^2}}}{2}$$ .

attempt to substitute ($$0$$, $$10$$) into their integrated expression (even if $$C$$ is missing)     (M1)

correct working     (A1)

eg   $$10 = 3 + C$$ , $$C = 7$$

$$s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7$$     A1     N6

Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of $${\rm{e}}$$.

[7 marks]

Question

A toy car travels with velocity v ms−1 for six seconds. This is shown in the graph below.

The following diagram shows the graph of $$y = f(x)$$, for $$– 4 \le x \le 5$$.

Write down the car’s velocity at $$t = 3$$ .

[1]
a.

Write down the value of $$f( – 3)$$;

[1]
a(i).

Find the car’s acceleration at $$t = 1.5$$ .

[2]
b.

Find the total distance travelled.

[3]
c.
Answer/Explanation

Markscheme

$$4{\text{ (m}}{{\text{s}}^{ – 1}}{\text{)}}$$     A1     N1

[1 mark]

a.

$$f( – 3) = – 1$$     A1     N1

[1 mark]

a(i).

recognizing that acceleration is the gradient     M1

e.g. $$a(1.5) = \frac{{4 – 0}}{{2 – 0}}$$

$$a = 2$$ $${\text{(m}}{{\text{s}}^{ – 2}}{\text{)}}$$     A1     N1

[2 marks]

b.

recognizing area under curve     M1

e.g. trapezium, triangles, integration

correct substitution     A1

e.g. $$\frac{1}{2}(3 + 6)4$$ , $$\int_0^6 {\left| {v(t)} \right|} {\rm{d}}t$$

distance 18 (m)     A1     N2

[3 marks]

c.

Question

The acceleration, $$a{\text{ m}}{{\text{s}}^{ – 2}}$$, of a particle at time t seconds is given by $$a = 2t + \cos t$$ .

Find the acceleration of the particle at $$t = 0$$ .

[2]
a.

Find the velocity, v, at time t, given that the initial velocity of the particle is $${\text{m}}{{\text{s}}^{ – 1}}$$ .

[5]
b.

Find $$\int_0^3 {v{\rm{d}}t}$$ , giving your answer in the form $$p – q\cos 3$$ .

[7]
c.

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

[2]
d.
Answer/Explanation

Markscheme

substituting $$t = 0$$     (M1)

e.g. $$a(0) = 0 + \cos 0$$

$$a(0) = 1$$     A1     N2

[2 marks]

a.

evidence of integrating the acceleration function     (M1)

e.g. $$\int {(2t + \cos t){\text{d}}t}$$

correct expression $${t^2} + \sin t + c$$     A1A1

Note: If “$$+ c$$” is omitted, award no further marks.

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

e.g. $$2 = 0 + \sin 0 + c$$ , $$c = 2$$

$$v(t) = {t^2} + \sin t + 2$$     A1     N3

[5 marks]

b.

$$\int {({t^2} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} – \cos t + 2t$$     A1A1A1

Note: Award A1 for each correct term.

evidence of using $$v(3) – v(0)$$     (M1)

correct substitution     A1

e.g. $$(9 – \cos 3 + 6) – (0 – \cos 0 + 0)$$ , $$(15 – \cos 3) – ( – 1)$$

$$16 – \cos 3$$ (accept $$p = 16$$ , $$q = – 1$$ )     A1A1     N3

[7 marks]

c.

reference to motion, reference to first 3 seconds     R1R1     N2

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

[2 marks]

d.

Question

The velocity v ms−1 of a particle at time t seconds, is given by $$v = 2t + \cos 2t$$ , for $$0 \le t \le 2$$ .

Write down the velocity of the particle when $$t = 0$$ .

[1]
a.

When $$t = k$$ , the acceleration is zero.

(i)     Show that $$k = \frac{\pi }{4}$$ .

(ii)    Find the exact velocity when $$t = \frac{\pi }{4}$$ .

[8]
b(i) and (ii).

When $$t < \frac{\pi }{4}$$ , $$\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0$$ and when $$t > \frac{\pi }{4}$$ , $$\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0$$  .

Sketch a graph of v against t .

[4]
c.

Let d be the distance travelled by the particle for $$0 \le t \le 1$$ .

(i)     Write down an expression for d .

(ii)    Represent d on your sketch.

[3]
d(i) and (ii).
Answer/Explanation

Markscheme

$$v = 1$$     A1     N1

[1 mark]

a.

(i) $$\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2$$     A1

$$\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t$$     A1A1

Note: Award A1 for coefficient 2 and A1 for $$– \sin 2t$$ .

evidence of considering acceleration = 0     (M1)

e.g. $$\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0$$ , $$2 – 2\sin 2t = 0$$

correct manipulation     A1

e.g. $$\sin 2k = 1$$ , $$\sin 2t = 1$$

$$2k = \frac{\pi }{2}$$ (accept $$2t = \frac{\pi }{2}$$ )     A1

$$k = \frac{\pi }{4}$$     AG     N0

(ii) attempt to substitute $$t = \frac{\pi }{4}$$ into v     (M1)

e.g. $$2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)$$

$$v = \frac{\pi }{2}$$     A1     N2

[8 marks]

b(i) and (ii).

A1A1A2      N4

Notes: Award A1 for y-intercept at $$(0{\text{, }}1)$$ , A1 for curve having zero gradient at $$t = \frac{\pi }{4}$$ , A2 for shape that is concave down to the left of $$\frac{\pi }{4}$$ and concave up to the right of $$\frac{\pi }{4}$$ . If a correct curve is drawn without indicating $$t = \frac{\pi }{4}$$ , 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 essential features need to be clear.

[4 marks]

c.

(i) correct expression     A2

e.g. $$\int_0^1 {(2t + \cos 2t){\rm{d}}t}$$ , $$\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1$$ , $$1 + \frac{{\sin 2}}{2}$$ , $$\int_0^1 {v{\rm{d}}t}$$

(ii)

A1

Note: The line at $$t = 1$$ needs to be clearly after $$t = \frac{\pi }{4}$$ .

[3 marks]

d(i) and (ii).

Question

In this question, you are given that $$\cos \frac{\pi }{3} = \frac{1}{2}$$ , and $$\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}$$ .

The displacement of an object from a fixed point, O is given by $$s(t) = t – \sin 2t$$ for $$0 \le t \le \pi$$ .

Find $$s'(t)$$ .

[3]
a.

In this interval, there are only two values of t for which the object is not moving. One value is $$t = \frac{\pi }{6}$$ .

Find the other value.

[4]
b.

Show that $$s'(t) > 0$$ between these two values of t .

[3]
c.

Find the distance travelled between these two values of t .

[5]
d.
Answer/Explanation

Markscheme

$$s'(t) = 1 – 2\cos 2t$$    A1A2     N3

Note: Award A1 for 1, A2 for $$- 2\cos 2t$$ .

[3 marks]

a.

evidence of valid approach     (M1)

e.g. setting $$s'(t) = 0$$

correct working     A1

e.g. $$2\cos 2t = 1$$ , $$\cos 2t = \frac{1}{2}$$

$$2t = \frac{\pi }{3}$$ , $$\frac{{5\pi }}{3}$$ , $$\ldots$$     (A1)

$$t = \frac{{5\pi }}{6}$$     A1     N3

[4 marks]

b.

evidence of valid approach     (M1)

e.g. choosing a value in the interval $$\frac{\pi }{6} < t < \frac{{5\pi }}{6}$$

correct substitution     A1

e.g. $$s’\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi$$

$$s’\left( {\frac{\pi }{2}} \right) = 3$$     A1

$$s'(t) > 0$$     AG     N0

[3 marks]

c.

evidence of approach using s or integral of $$s’$$     (M1)

e.g. $$\int {s'(t){\rm{d}}t}$$ ; $$s\left( {\frac{{5\pi }}{6}} \right)$$ , $$s\left( {\frac{\pi }{6}} \right)$$ ; $$\left[ {t – \sin 2t} \right]_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}}$$

substituting values and subtracting     (M1)

e.g. $$s\left( {\frac{{5\pi }}{6}} \right) – s\left( {\frac{\pi }{6}} \right)$$ , $$\left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right) – \left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right)$$

correct substitution     A1

e.g. $$\frac{{5\pi }}{6} – \sin \frac{{5\pi }}{3} – \left[ {\frac{\pi }{6} – \sin \frac{\pi }{3}} \right]$$ , $$\left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right) – \left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right)$$

distance is $$\frac{{2\pi }}{3} + \sqrt 3$$     A1A1     N3

Note: Award A1 for $$\frac{{2\pi }}{3}$$ , A1 for $$\sqrt 3$$ .

[5 marks]

d.