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 .
Find an expression for the velocity of P1 at time t . [2]
Particle P2 also moves in a straight line. The position of P2Â is given by
The speed of P1 is greater than the speed of P2 when t > q .
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. \(s = \int {(6{{\rm{e}}^{3x}} + 4)} {\rm{d}}x\)
\(s = 2{{\rm{e}}^{3t}} + 4t + C\)Â Â Â Â Â A2A1
substituting \(t = 0\) ,   (M1)
\(7 = 2 + C\)Â Â Â Â A1
\(C = 5\)
\(s = 2{{\rm{e}}^{3t}} + 4t + 5\)Â Â Â A1 Â Â 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 \(t = 0\) .
Find the velocity, v, at time t, given that the initial velocity of the particle is \({\text{m}}{{\text{s}}^{ – 1}}\) .
Find \(\int_0^3 {v{\rm{d}}t} \) , giving your answer in the form \(p – q\cos 3\) .
What information does the answer to part (c) give about the motion of the particle?
Answer/Explanation
Markscheme
substituting \(t = 0\)Â Â Â Â (M1)
e.g. \(a(0) = 0 + \cos 0\)
\(a(0) = 1\)Â Â Â A1 Â Â N2
[2 marks]
evidence of integrating the acceleration function    (M1)
e.g. \(\int {(2t + \cos t){\text{d}}t} \)
correct expression \({t^2} + \sin t + c\)Â Â Â Â A1A1
Note: If “\( + c\)” is omitted, award no further marks.
evidence of substituting (2,0) into indefinite integral    (M1)
e.g. \(2 = 0 + \sin 0 + c\) , \(c = 2\)
\(v(t) = {t^2} + \sin t + 2\)Â Â Â A1 Â Â N3
[5 marks]
\(\int {({t^2} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} – \cos t + 2t\)Â Â Â A1A1A1
Note: Award A1 for each correct term.
evidence of using \(v(3) – v(0)\)Â Â Â Â (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]
reference to motion, reference to first 3 seconds    R1R1    N2
e.g. displacement in 3 seconds, distance travelled in 3 seconds
[2 marks]
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.
Write down the value of t when the velocity is greatest.
Answer/Explanation
Markscheme
    A2A2    N4
[4 marks]
\(t = 3\)Â Â Â Â A2Â Â Â Â N2
[2 marks]
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)Â Â Â Â If \(s = 100\) when \(t = 0\) , find an expression for s in terms of a and t.
(ii)Â Â Â If \(s = 0\) when \(t = 0\) , write down an expression for s in terms of a and t.
A train M slows down so that it comes to a stop at the station.
(i)Â Â Â Â Find the time it takes train M to come to a stop, giving your answer in terms of a.
(ii)Â Â Â Hence show that \(a = \frac{8}{5}\) .
For a different train N, the value of a is 4.
Show that this train will stop before it reaches the station.
Answer/Explanation
Markscheme
Note: In this question, do not penalize absence of units.
(i) \(s = \int {(40 – at){\rm{d}}t} \)Â Â Â Â Â (M1)
\(s = 40t – \frac{1}{2}a{t^2} + c\)Â Â Â Â (A1)(A1)
substituting \(s = 100\)Â when \(t = 0\) (\(c = 100\) )Â Â Â Â Â (M1)
\(s = 40t – \frac{1}{2}a{t^2} + 100\)Â Â Â Â A1Â Â Â Â N5
(ii) \(s = 40t – \frac{1}{2}a{t^2}\)Â Â Â Â Â A1Â Â Â Â N1
[6 marks]
(i) stops at station, so \(v = 0\)Â Â Â Â Â (M1)
\(t = \frac{{40}}{a}\) (seconds)Â Â Â Â A1Â Â Â Â N2
(ii) evidence of choosing formula for s from (a) (ii)Â Â Â Â (M1)
substituting \(t = \frac{{40}}{a}\)Â Â Â Â (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 \(a = \frac{8}{5}\)Â Â Â Â Â A1
e.g. \(500a = 800\) , \(5 = \frac{8}{a}\) , \(1000a = 3200 – 1600\)
\(a = \frac{8}{5}\)Â Â Â Â AGÂ Â Â Â N0
[6 marks]
METHOD 1
\(v = 40 – 4t\) , stops when \(v = 0\)
\(40 – 4t = 0\)Â Â Â Â (A1)
\(t = 10\)Â Â Â Â A1
substituting into expression for s    M1
\(s = 40 \times 10 – \frac{1}{2} \times 4 \times {10^2}\)
\(s = 200\)Â Â Â Â A1
since \(200 < 500\) (allow FT on their s, if \(s < 500\)Â )Â Â Â Â R1
train stops before the station    AG   N0
METHOD 2
from (b) \(t = \frac{{40}}{4} = 10\)Â Â Â Â Â A2
substituting into expression for s
e.g. \(s = 40 \times 10 – \frac{1}{2} \times 4 \times {10^2}\)Â Â Â Â Â M1
\(s = 200\)Â Â Â Â A1
since \(200 < 500\)Â Â Â Â Â R1
train stops before the station    AG    N0
METHOD 3
a is deceleration    A2
\(4 > \frac{8}{5}\)Â Â Â A1
so stops in shorter time    (A1)
so less distance travelled    R1
so stops before station    AG    N0
[5 marks]
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 \(t = 0\) .
When \(t = k\) , the acceleration is zero.
(i)Â Â Â Â Show that \(k = \frac{\pi }{4}\)Â .
(ii) Â Â Find the exact velocity when \(t = \frac{\pi }{4}\)Â .
When \(t < \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)Â and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)Â Â .
Sketch a graph of v against t .
Let d be the distance travelled by the particle for \(0 \le t \le 1\) .
(i)Â Â Â Â Write down an expression for d .
(ii)Â Â Â Represent d on your sketch.
Answer/Explanation
Markscheme
\(v = 1\)Â Â Â Â A1Â Â Â Â N1
[1 mark]
(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\)Â Â Â Â Â A1
\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t\)Â Â Â Â A1A1
Note: Award A1 for coefficient 2 and A1 for \( – \sin 2t\) .
evidence of considering acceleration = 0Â Â Â Â (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
\(k = \frac{\pi }{4}\)Â Â Â Â AGÂ Â Â Â N0
(ii) attempt to substitute \(t = \frac{\pi }{4}\) into v    (M1)
e.g. \(2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)\)
\(v = \frac{\pi }{2}\)Â Â Â Â A1Â Â Â Â N2
[8 marks]
    A1A1A2     N4
Notes: Award A1 for y-intercept at \((0{\text{, }}1)\) , A1 for curve having zero gradient at \(t = \frac{\pi }{4}\)Â , A2 for shape that is concave down to the left of \(\frac{\pi }{4}\)Â and concave up to the right of \(\frac{\pi }{4}\)Â . If a correct curve is drawn without indicating \(t = \frac{\pi }{4}\)Â , 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]
(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 \(t = \frac{\pi }{4}\)Â .
[3 marks]
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 \(f(x) = – 10(x – p)(x – q)\) .
Find another expression for \(f(x)\) in the form \(f(x) = – 10{(x – h)^2} + k\) .
Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x – 10{x^2}\) .
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) Â Â Find the value of t when the speed of the particle is greatest.
(ii) Â Â Find the acceleration of the particle when its speed is zero.
Answer/Explanation
Markscheme
\(f(x) = – 10(x + 4)(x – 6)\)Â Â Â Â A1A1 Â Â N2
[2 marks]
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. \(k = – 10(1 + 4)(1 – 6)\)
\(f(x) = – 10{(x – 1)^2} + 250\)Â Â Â Â Â A1A1 Â Â N4
METHOD 2
attempt to expand \(f(x)\)Â Â Â Â Â (M1)
e.g. \( – 10({x^2} – 2x – 24)\)
attempt to complete the square    (M1)
e.g. \( – 10({(x – 1)^2} – 1 – 24)\)
\(f(x) = – 10{(x – 1)^2} + 250\)Â Â Â Â A1A1Â Â Â Â N4
[4 marks]
attempt to simplify    (M1)
e.g. distributive property, \( – 10(x – 1)(x – 1) + 250\)
correct simplification    A1
e.g. \( – 10({x^2} – 6x + 4x – 24)\) , \( – 10({x^2} – 2x + 1) + 250\)
\(f(x) = 240 + 20x – 10{x^2}\)Â Â Â Â Â AGÂ Â Â Â N0
[2 marks]
(i) valid approach    (M1)
e.g. vertex of parabola, \(v'(t) = 0\)
\(t = 1\)Â Â Â Â Â A1Â Â Â Â N2
(ii) recognizing \(a(t) = v'(t)\)Â Â Â Â Â (M1)
\(a(t) = 20 – 20t\)Â Â Â Â Â A1A1
speed is zero \( \Rightarrow t = 6\)Â Â Â Â Â (A1)
\(a(6) = – 100\) (\({\text{m}}{{\text{s}}^{ – 2}}\))Â Â Â Â A1Â Â Â Â N3
[7 marks]
Question
In this question, you are given that \(\cos \frac{\pi }{3} = \frac{1}{2}\)Â , and \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\)Â .
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 \(s'(t)\) .
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.
Show that \(s'(t) > 0\) between these two values of t .
Find the distance travelled between these two values of t .
Answer/Explanation
Markscheme
\(s'(t) = 1 – 2\cos 2t\)Â Â Â A1A2 Â Â N3
Note: Award A1 for 1, A2 for \(- 2\cos 2t\)Â .
[3 marks]
evidence of valid approach    (M1)
e.g. setting \(s'(t) = 0\)
correct working    A1
e.g. \(2\cos 2t = 1\) , \(\cos 2t = \frac{1}{2}\)
\(2t = \frac{\pi }{3}\) , \(\frac{{5\pi }}{3}\) , \(\ldots \)Â Â Â Â (A1)
\(t = \frac{{5\pi }}{6}\)Â Â Â Â A1 Â Â N3Â
[4 marks]
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. \(s’\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi \)
\(s’\left( {\frac{\pi }{2}} \right) = 3\)Â Â Â Â A1
\(s'(t) > 0\)Â Â Â Â AG Â Â N0
[3 marks]
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 \(\frac{{2\pi }}{3} + \sqrt 3 \)Â Â Â Â Â A1A1Â Â Â Â N3
Note: Award A1 for \(\frac{{2\pi }}{3}\) , A1 for \(\sqrt 3 \) .
[5 marks]
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  \(\int {(6{{\rm{e}}^{2t}} + t)} \)
\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + C\)Â Â Â Â A2A1
Note: Award A2 for \(3{{\rm{e}}^{2t}}\)Â , A1 for \(\frac{{{t^2}}}{2}\)Â .
attempt to substitute (\(0\), \(10\))Â into their integrated expression (even if \(C\) is missing)Â Â Â Â (M1)Â
correct working    (A1)
eg  \(10 = 3 + C\) , \(C = 7\)
\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7\)Â Â Â Â A1Â Â Â Â 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 \(t = 3\) .
Write down the value of \(f( – 3)\);
Find the car’s acceleration at \(t = 1.5\) .
Find the total distance travelled.
Answer/Explanation
Markscheme
\(4{\text{ (m}}{{\text{s}}^{ – 1}}{\text{)}}\) Â Â Â A1Â Â Â Â N1
[1 mark]
\(f( – 3) =Â – 1\) Â Â A1 Â Â N1
[1 mark]
recognizing that acceleration is the gradient    M1
e.g. \(a(1.5) = \frac{{4 – 0}}{{2 – 0}}\)
\(a = 2\) \({\text{(m}}{{\text{s}}^{ – 2}}{\text{)}}\) Â Â Â A1 Â Â N1
[2 marks]
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)Â Â Â Â A1 Â Â N2
[3 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 \(t = 0\) .
Find the velocity, v, at time t, given that the initial velocity of the particle is \({\text{m}}{{\text{s}}^{ – 1}}\) .
Find \(\int_0^3 {v{\rm{d}}t} \) , giving your answer in the form \(p – q\cos 3\) .
What information does the answer to part (c) give about the motion of the particle?
Answer/Explanation
Markscheme
substituting \(t = 0\)Â Â Â Â (M1)
e.g. \(a(0) = 0 + \cos 0\)
\(a(0) = 1\)Â Â Â A1 Â Â N2
[2 marks]
evidence of integrating the acceleration function    (M1)
e.g. \(\int {(2t + \cos t){\text{d}}t} \)
correct expression \({t^2} + \sin t + c\)Â Â Â Â A1A1
Note: If “\( + c\)” is omitted, award no further marks.
evidence of substituting (2,0) into indefinite integral    (M1)
e.g. \(2 = 0 + \sin 0 + c\) , \(c = 2\)
\(v(t) = {t^2} + \sin t + 2\)Â Â Â A1 Â Â N3
[5 marks]
\(\int {({t^2} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} – \cos t + 2t\)Â Â Â A1A1A1
Note: Award A1 for each correct term.
evidence of using \(v(3) – v(0)\)Â Â Â Â (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]
reference to motion, reference to first 3 seconds    R1R1    N2
e.g. displacement in 3 seconds, distance travelled in 3 seconds
[2 marks]
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 \(t = 0\) .
When \(t = k\) , the acceleration is zero.
(i)Â Â Â Â Show that \(k = \frac{\pi }{4}\)Â .
(ii) Â Â Find the exact velocity when \(t = \frac{\pi }{4}\)Â .
When \(t < \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)Â and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)Â Â .
Sketch a graph of v against t .
Let d be the distance travelled by the particle for \(0 \le t \le 1\) .
(i)Â Â Â Â Write down an expression for d .
(ii)Â Â Â Represent d on your sketch.
Answer/Explanation
Markscheme
\(v = 1\)Â Â Â Â A1Â Â Â Â N1
[1 mark]
(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\)Â Â Â Â Â A1
\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t\)Â Â Â Â A1A1
Note: Award A1 for coefficient 2 and A1 for \( – \sin 2t\) .
evidence of considering acceleration = 0Â Â Â Â (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
\(k = \frac{\pi }{4}\)Â Â Â Â AGÂ Â Â Â N0
(ii) attempt to substitute \(t = \frac{\pi }{4}\) into v    (M1)
e.g. \(2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)\)
\(v = \frac{\pi }{2}\)Â Â Â Â A1Â Â Â Â N2
[8 marks]
    A1A1A2     N4
Notes: Award A1 for y-intercept at \((0{\text{, }}1)\) , A1 for curve having zero gradient at \(t = \frac{\pi }{4}\)Â , A2 for shape that is concave down to the left of \(\frac{\pi }{4}\)Â and concave up to the right of \(\frac{\pi }{4}\)Â . If a correct curve is drawn without indicating \(t = \frac{\pi }{4}\)Â , 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]
(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 \(t = \frac{\pi }{4}\)Â .
[3 marks]
Question
In this question, you are given that \(\cos \frac{\pi }{3} = \frac{1}{2}\)Â , and \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\)Â .
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 \(s'(t)\) .
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.
Show that \(s'(t) > 0\) between these two values of t .
Find the distance travelled between these two values of t .
Answer/Explanation
Markscheme
\(s'(t) = 1 – 2\cos 2t\)Â Â Â A1A2 Â Â N3
Note: Award A1 for 1, A2 for \(- 2\cos 2t\)Â .
[3 marks]
evidence of valid approach    (M1)
e.g. setting \(s'(t) = 0\)
correct working    A1
e.g. \(2\cos 2t = 1\) , \(\cos 2t = \frac{1}{2}\)
\(2t = \frac{\pi }{3}\) , \(\frac{{5\pi }}{3}\) , \(\ldots \)Â Â Â Â (A1)
\(t = \frac{{5\pi }}{6}\)Â Â Â Â A1 Â Â N3Â
[4 marks]
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. \(s’\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi \)
\(s’\left( {\frac{\pi }{2}} \right) = 3\)Â Â Â Â A1
\(s'(t) > 0\)Â Â Â Â AG Â Â N0
[3 marks]
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 \(\frac{{2\pi }}{3} + \sqrt 3 \)Â Â Â Â Â A1A1Â Â Â Â N3
Note: Award A1 for \(\frac{{2\pi }}{3}\) , A1 for \(\sqrt 3 \) .
[5 marks]