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.

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