what information does the answer to part c give about the motion of the particle
Question
A particle moves along a direct line so that its velocity, \(v{\text{ g}}{{\text{southward}}^{ – 1}}\)at time t seconds is given past \(v = 6{{\rm{due east}}^{3t}} + four\) . When \(t = 0\) , the deportation, s, of the particle is 7 metres. Find an expression for s in terms of t.
Answer/Explanation
Markscheme
prove of anti-differentiation (M1)
eastward.thousand. \(south = \int {(6{{\rm{e}}^{3x}} + 4)} {\rm{d}}10\)
\(south = two{{\rm{e}}^{3t}} + 4t + C\) A2A1
substituting \(t = 0\) , (M1)
\(7 = 2 + C\) A1
\(C = 5\)
\(s = ii{{\rm{e}}^{3t}} + 4t + 5\) A1 N3
[7 marks]
Question
The acceleration, \(a{\text{ m}}{{\text{s}}^{ – two}}\), of a particle at time t seconds is given by \(a = 2t + \cos t\) .
Observe the acceleration of the particle at \(t = 0\) .
[2]
a.
Find the velocity, five, at time t, given that the initial velocity of the particle is \({\text{m}}{{\text{s}}^{ – 1}}\) .
[5]
b.
Find \(\int_0^3 {5{\rm{d}}t} \) , giving your answer in the class \(p – q\cos three\) .
[7]
c.
What information does the reply to part (c) give about the movement of the particle?
[two]
d.
Reply/Explanation
Markscheme
substituting \(t = 0\) (M1)
due east.m. \(a(0) = 0 + \cos 0\)
\(a(0) = 1\) A1 N2
[two marks]
a.
show of integrating the acceleration function (M1)
e.grand. \(\int {(2t + \cos t){\text{d}}t} \)
correct expression \({t^2} + \sin t + c\) A1A1
Note: If "\( + c\)" is omitted, award no further marks.
bear witness of substituting (ii,0) into indefinite integral (M1)
e.k. \(2 = 0 + \sin 0 + c\) , \(c = two\)
\(five(t) = {t^2} + \sin t + two\) 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 \(v(iii) – 5(0)\) (M1)
correct substitution A1
eastward.one thousand. \((9 – \cos 3 + 6) – (0 – \cos 0 + 0)\) , \((15 – \cos iii) – ( – one)\)
\(16 – \cos 3\) (take \(p = 16\) , \(q = – 1\) ) A1A1 N3
[7 marks]
c.
reference to motion, reference to first 3 seconds R1R1 N2
e.g. displacement in three seconds, distance travelled in three seconds
[ii marks]
d.
Question
The following diagram shows the graphs of the displacement, velocity and acceleration of a moving object equally functions of time, t.
Complete the following tabular array past 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
[iv marks]
a.
\(t = 3\) A2 N2
[2 marks]
b.
Question
In this question due south represents displacement in metres and t represents time in seconds.
The velocity v chiliad s–1 of a moving torso is given past \(5 = 40 – at\) where a is a not-zilch constant.
Trains approaching a station start to slow down when they pass a point P. As a railroad train slows down, its velocity is given past \(v = 40 – at\) , where \(t = 0\) at P. The station is 500 m from P.
(i) If \(southward = 100\) when \(t = 0\) , discover an expression for s in terms of a and t.
(two) If \(due south = 0\) when \(t = 0\) , write downwardly an expression for south in terms of a and t.
[6]
a.
A train M slows down so that it comes to a finish at the station.
(i) Notice the time it takes train M to come to a end, giving your answer in terms of a.
(ii) Hence prove that \(a = \frac{8}{5}\) .
[vi]
b.
For a different train Northward, the value of a is 4.
Show that this train will stop before it reaches the station.
[5]
c.
Reply/Explanation
Markscheme
Annotation: In this question, do not penalize absence of units.
(i) \(s = \int {(twoscore – at){\rm{d}}t} \) (M1)
\(s = 40t – \frac{ane}{ii}a{t^ii} + c\) (A1)(A1)
substituting \(south = 100\) 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, and then \(v = 0\) (M1)
\(t = \frac{{xl}}{a}\) (seconds) A1 N2
(2) bear witness of choosing formula for s from (a) (ii) (M1)
substituting \(t = \frac{{40}}{a}\) (M1)
e.one thousand. \(40 \times \frac{{40}}{a} – \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^two}}}\)
setting up equation M1
e.g. \(500 = s\) , \(500 = 40 \times \frac{{40}}{a} – \frac{1}{two}a \times \frac{{{{40}^2}}}{{{a^2}}}\) , \(500 = \frac{{1600}}{a} – \frac{{800}}{a}\)
evidence of simplification to an expression which obviously leads to \(a = \frac{eight}{v}\) A1
e.g. \(500a = 800\) , \(5 = \frac{8}{a}\) , \(1000a = 3200 – 1600\)
\(a = \frac{8}{5}\) AG N0
[6 marks]
b.
METHOD 1
\(five = 40 – 4t\) , stops when \(v = 0\)
\(xl – 4t = 0\) (A1)
\(t = 10\) A1
substituting into expression for s M1
\(south = 40 \times 10 – \frac{1}{2} \times four \times {10^2}\)
\(s = 200\) A1
since \(200 < 500\) (allow FT on their s, if \(s < 500\) ) R1
railroad train stops before the station AG N0
METHOD 2
from (b) \(t = \frac{{40}}{4} = x\) A2
substituting into expression for south
e.g. \(southward = xl \times 10 – \frac{1}{2} \times 4 \times {x^ii}\) M1
\(southward = 200\) A1
since \(200 < 500\) R1
railroad 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
[v 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 \(t = 0\) .
[ane]
a.
When \(t = k\) , the acceleration is zip.
(i) Prove that \(chiliad = \frac{\pi }{4}\) .
(ii) Find the exact velocity when \(t = \frac{\pi }{4}\) .
[8]
b(i) and (2).
When \(t < \frac{\pi }{iv}\) , \(\frac{{{\rm{d}}five}}{{{\rm{d}}t}} > 0\) and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\) .
Sketch a graph of v against t .
[iv]
c.
Permit d be the altitude travelled by the particle for \(0 \le t \le i\) .
(i) Write down an expression for d .
(ii) Represent d on your sketch.
[3]
d(i) and (2).
Answer/Explanation
Markscheme
\(five = 1\) A1 N1
[one marker]
a.
(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\) A1
\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t\) A1A1
Note: Honour A1 for coefficient ii and A1 for \( – \sin 2t\) .
evidence of because acceleration = 0 (M1)
e.g. \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0\) , \(two – 2\sin 2t = 0\)
correct manipulation A1
eastward.g. \(\sin 2k = 1\) , \(\sin 2t = ane\)
\(2k = \frac{\pi }{2}\) (accept \(2t = \frac{\pi }{2}\) ) A1
\(k = \frac{\pi }{4}\) AG N0
(ii) attempt to substitute \(t = \frac{\pi }{iv}\) into v (M1)
due east.g. \(ii\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{ii\pi }}{four}} \right)\)
\(v = \frac{\pi }{ii}\) A1 N2
[eight marks]
b(i) and (2).
A1A1A2 N4
Notes: Award A1 for y-intercept at \((0{\text{, }}ane)\) , A1 for curve having naught gradient at \(t = \frac{\pi }{four}\) , A2 for shape that is concave downward to the left of \(\frac{\pi }{iv}\) and concave up to the right of \(\frac{\pi }{four}\) . If a correct curve is drawn without indicating \(t = \frac{\pi }{iv}\) , do not award the 2d A1 for the zero gradient, but honour the last A2 if appropriate. Sketch demand not be drawn to scale. Merely essential features need to be clear.
[4 marks]
c.
(i) correct expression A2
e.g. \(\int_0^1 {(2t + \cos 2t){\rm{d}}t} \) , \(\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1\) , \(1 + \frac{{\sin 2}}{2}\) , \(\int_0^one {v{\rm{d}}t} \)
(ii)
A1
Note: The line at \(t = one\) needs to exist conspicuously subsequently \(t = \frac{\pi }{4}\) .
[3 marks]
d(i) and (ii).
Question
The post-obit diagram shows role 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 course \(f(x) = – 10(ten – p)(x – q)\) .
[ii]
a.
Detect another expression for \(f(x)\) in the grade \(f(x) = – 10{(x – h)^2} + k\) .
[4]
b.
Bear witness that \(f(x)\) can also be written in the course \(f(x) = 240 + 20x – 10{10^2}\) .
[2]
c.
A particle moves forth a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ – ane}}\) , 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) Notice the acceleration of the particle when its speed is nil.
[7]
d(i) and (ii).
Answer/Explanation
Markscheme
\(f(10) = – 10(x + 4)(x – 6)\) A1A1 N2
[two marks]
a.
METHOD ane
attempting to find the 10-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)
east.k. \(thou = – ten(1 + 4)(ane – 6)\)
\(f(x) = – 10{(ten – one)^2} + 250\) A1A1 N4
METHOD ii
attempt to expand \(f(10)\) (M1)
e.one thousand. \( – 10({x^ii} – 2x – 24)\)
attempt to complete the square (M1)
e.g. \( – ten({(x – 1)^ii} – 1 – 24)\)
\(f(x) = – 10{(x – 1)^ii} + 250\) A1A1 N4
[4 marks]
b.
attempt to simplify (M1)
e.g. distributive property, \( – x(x – 1)(x – 1) + 250\)
right simplification A1
e.thou. \( – 10({x^ii} – 6x + 4x – 24)\) , \( – 10({x^ii} – 2x + 1) + 250\)
\(f(ten) = 240 + 20x – 10{10^2}\) AG N0
[ii marks]
c.
(i) valid arroyo (M1)
e.g. vertex of parabola, \(five'(t) = 0\)
\(t = 1\) A1 N2
(two) recognizing \(a(t) = v'(t)\) (M1)
\(a(t) = 20 – 20t\) A1A1
speed is null \( \Rightarrow t = 6\) (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 \(\cos \frac{\pi }{three} = \frac{i}{2}\) , and \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{ii}\) .
The deportation of an object from a fixed point, O is given past \(due south(t) = t – \sin 2t\) for \(0 \le t \le \pi \) .
In this interval, there are simply two values of t for which the object is non moving. One value is \(t = \frac{\pi }{6}\) .
Find the other value.
[4]
b.
Show that \(south'(t) > 0\) betwixt these ii values of t .
[three]
c.
Find the distance travelled between these two values of t .
[5]
d.
Respond/Explanation
Markscheme
\(s'(t) = 1 – ii\cos 2t\) A1A2 N3
Note: Honor A1 for 1, A2 for \(- 2\cos 2t\) .
[3 marks]
a.
evidence of valid approach (M1)
e.g. setting \(due south'(t) = 0\)
right working A1
e.g. \(ii\cos 2t = 1\) , \(\cos 2t = \frac{1}{2}\)
\(2t = \frac{\pi }{three}\) , \(\frac{{5\pi }}{3}\) , \(\ldots \) (A1)
\(t = \frac{{5\pi }}{half dozen}\) A1 N3
[4 marks]
b.
bear witness of valid approach (M1)
due east.grand. choosing a value in the interval \(\frac{\pi }{6} < t < \frac{{five\pi }}{half-dozen}\)
right substitution A1
e.m. \(s'\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi \)
\(s'\left( {\frac{\pi }{2}} \right) = 3\) A1
\(south'(t) > 0\) AG N0
[iii marks]
c.
show of approach using s or integral of \(s'\) (M1)
east.grand. \(\int {south'(t){\rm{d}}t} \) ; \(s\left( {\frac{{5\pi }}{vi}} \correct)\) , \(s\left( {\frac{\pi }{half-dozen}} \right)\) ; \(\left[ {t – \sin 2t} \right]_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}}\)
substituting values and subtracting (M1)
e.chiliad. \(southward\left( {\frac{{v\pi }}{6}} \right) – due south\left( {\frac{\pi }{six}} \right)\) , \(\left( {\frac{\pi }{half-dozen} – \frac{{\sqrt three }}{2}} \right) – \left( {\frac{{v\pi }}{vi} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right)\)
correct substitution A1
east.g. \(\frac{{v\pi }}{half dozen} – \sin \frac{{5\pi }}{iii} – \left[ {\frac{\pi }{6} – \sin \frac{\pi }{iii}} \right]\) , \(\left( {\frac{{five\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right) – \left( {\frac{\pi }{half-dozen} – \frac{{\sqrt 3 }}{2}} \right)\)
distance is \(\frac{{two\pi }}{three} + \sqrt three \) A1A1 N3
Note: Award A1 for \(\frac{{2\pi }}{iii}\) , A1 for \(\sqrt three \) .
[5 marks]
d.
Question
A rocket moving in a straight line has velocity \(v\) km s–1 and deportation \(s\) km at time \(t\) seconds. The velocity \(v\) is given by \(5(t) = half dozen{{\rm{e}}^{2t}} + t\) . When \(t = 0\) , \(south = x\) .
Find an expression for the displacement of the rocket in terms of \(t\) .
Respond/Explanation
Markscheme
testify of anti-differentiation (M1)
eg \(\int {(6{{\rm{e}}^{2t}} + t)} \)
\(s = 3{{\rm{eastward}}^{2t}} + \frac{{{t^2}}}{2} + C\) A2A1
Note: Honour A2 for \(3{{\rm{e}}^{2t}}\) , A1 for \(\frac{{{t^two}}}{2}\) .
endeavor to substitute (\(0\), \(10\)) into their integrated expression (even if \(C\) is missing) (M1)
correct working (A1)
eg \(ten = 3 + C\) , \(C = 7\)
\(s = three{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7\) A1 N6
Note: Exception to the FT rule. If working shown, allow full FT on wrong integration which must involve a ability 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 post-obit diagram shows the graph of \(y = f(x)\), for \( – 4 \le x \le 5\).
Write down the car'south velocity at \(t = three\) .
[1]
a.
Write down the value of \(f( – 3)\);
[ane]
a(i).
Find the motorcar's acceleration at \(t = ane.5\) .
[two]
b.
Find the total distance travelled.
[3]
c.
Answer/Explanation
Markscheme
\(4{\text{ (m}}{{\text{s}}^{ – 1}}{\text{)}}\) A1 N1
[1 mark]
a.
\(f( – three) = – one\) A1 N1
[one mark]
a(i).
recognizing that acceleration is the gradient M1
eastward.thousand. \(a(ane.five) = \frac{{4 – 0}}{{2 – 0}}\)
\(a = ii\) \({\text{(thousand}}{{\text{s}}^{ – two}}{\text{)}}\) A1 N1
[two marks]
b.
recognizing surface area under bend M1
east.g. trapezium, triangles, integration
right commutation A1
e.g. \(\frac{1}{2}(iii + 6)4\) , \(\int_0^6 {\left| {v(t)} \right|} {\rm{d}}t\)
distance 18 (one thousand) A1 N2
[3 marks]
c.
Question
The acceleration, \(a{\text{ m}}{{\text{s}}^{ – 2}}\), of a particle at fourth dimension t seconds is given by \(a = 2t + \cos t\) .
Find the acceleration of the particle at \(t = 0\) .
[2]
a.
Detect the velocity, v, at fourth dimension t, given that the initial velocity of the particle is \({\text{grand}}{{\text{southward}}^{ – 1}}\) .
[5]
b.
Find \(\int_0^3 {v{\rm{d}}t} \) , giving your reply in the class \(p – q\cos 3\) .
[7]
c.
What information does the reply to office (c) requite about the motion of the particle?
[two]
d.
Respond/Caption
Markscheme
substituting \(t = 0\) (M1)
e.g. \(a(0) = 0 + \cos 0\)
\(a(0) = 1\) A1 N2
[2 marks]
a.
bear witness of integrating the acceleration function (M1)
e.g. \(\int {(2t + \cos t){\text{d}}t} \)
correct expression \({t^2} + \sin t + c\) A1A1
Notation: If "\( + c\)" is omitted, laurels no further marks.
evidence of substituting (2,0) into indefinite integral (M1)
e.thou. \(ii = 0 + \sin 0 + c\) , \(c = 2\)
\(v(t) = {t^2} + \sin t + 2\) A1 N3
[five marks]
b.
\(\int {({t^two} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} – \cos t + 2t\) A1A1A1
Note: Award A1 for each correct term.
evidence of using \(v(3) – v(0)\) (M1)
correct substitution A1
e.yard. \((ix – \cos 3 + 6) – (0 – \cos 0 + 0)\) , \((15 – \cos 3) – ( – 1)\)
\(16 – \cos three\) (accept \(p = sixteen\) , \(q = – ane\) ) A1A1 N3
[7 marks]
c.
reference to move, reference to first 3 seconds R1R1 N2
e.thousand. displacement in 3 seconds, distance travelled in 3 seconds
[two marks]
d.
Question
The velocity 5 ms−1 of a particle at time t seconds, is given by \(five = 2t + \cos 2t\) , for \(0 \le t \le 2\) .
Write down the velocity of the particle when \(t = 0\) .
[i]
a.
When \(t = k\) , the acceleration is zero.
(i) Show that \(k = \frac{\pi }{4}\) .
(two) Notice the exact velocity when \(t = \frac{\pi }{4}\) .
[8]
b(i) and (2).
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 .
[four]
c.
Allow d exist the distance travelled by the particle for \(0 \le t \le 1\) .
(i) Write down an expression for d .
(two) Represent d on your sketch.
[3]
d(i) and (ii).
Answer/Explanation
Markscheme
\(five = one\) A1 N1
[ane 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 ii and A1 for \( – \sin 2t\) .
evidence of considering dispatch = 0 (M1)
e.thou. \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0\) , \(ii – 2\sin 2t = 0\)
correct manipulation A1
e.g. \(\sin 2k = one\) , \(\sin 2t = 1\)
\(2k = \frac{\pi }{ii}\) (accept \(2t = \frac{\pi }{two}\) ) A1
\(yard = \frac{\pi }{iv}\) AG N0
(2) attempt to substitute \(t = \frac{\pi }{4}\) into 5 (M1)
e.g. \(ii\left( {\frac{\pi }{four}} \right) + \cos \left( {\frac{{two\pi }}{four}} \correct)\)
\(v = \frac{\pi }{2}\) 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 slope at \(t = \frac{\pi }{4}\) , A2 for shape that is concave downward to the left of \(\frac{\pi }{4}\) and concave up to the correct of \(\frac{\pi }{4}\) . If a correct curve is drawn without indicating \(t = \frac{\pi }{iv}\) , exercise not award the second A1 for the zero gradient, simply honour the concluding A2 if appropriate. Sketch demand not exist drawn to scale. But essential features need to be clear.
[4 marks]
c.
(i) correct expression A2
eastward.g. \(\int_0^1 {(2t + \cos 2t){\rm{d}}t} \) , \(\left[ {{t^ii} + \frac{{\sin 2t}}{2}} \right]_0^ane\) , \(1 + \frac{{\sin 2}}{ii}\) , \(\int_0^1 {v{\rm{d}}t} \)
(ii)
A1
Note: The line at \(t = ane\) needs to be clearly after \(t = \frac{\pi }{4}\) .
[3 marks]
d(i) and (2).
Question
In this question, y'all are given that \(\cos \frac{\pi }{iii} = \frac{1}{two}\) , and \(\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{two}\) .
The displacement of an object from a fixed point, O is given past \(s(t) = t – \sin 2t\) for \(0 \le t \le \pi \) .
In this interval, at that place are simply 2 values of t for which the object is not moving. I 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 .
[five]
d.
Answer/Explanation
Markscheme
\(s'(t) = i – 2\cos 2t\) A1A2 N3
Annotation: Award A1 for i, A2 for \(- 2\cos 2t\) .
[3 marks]
a.
evidence of valid approach (M1)
e.g. setting \(due south'(t) = 0\)
correct working A1
east.g. \(2\cos 2t = ane\) , \(\cos 2t = \frac{1}{2}\)
\(2t = \frac{\pi }{3}\) , \(\frac{{five\pi }}{3}\) , \(\ldots \) (A1)
\(t = \frac{{5\pi }}{6}\) A1 N3
[4 marks]
b.
evidence of valid arroyo (M1)
e.g. choosing a value in the interval \(\frac{\pi }{6} < t < \frac{{v\pi }}{half-dozen}\)
correct substitution A1
e.g. \(south'\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi \)
\(s'\left( {\frac{\pi }{ii}} \right) = three\) A1
\(s'(t) > 0\) AG N0
[3 marks]
c.
evidence of approach using s or integral of \(southward'\) (M1)
e.g. \(\int {s'(t){\rm{d}}t} \) ; \(s\left( {\frac{{5\pi }}{6}} \right)\) , \(s\left( {\frac{\pi }{6}} \right)\) ; \(\left[ {t – \sin 2t} \correct]_{\frac{\pi }{six}}^{\frac{{5\pi }}{six}}\)
substituting values and subtracting (M1)
east.g. \(s\left( {\frac{{5\pi }}{6}} \right) – due south\left( {\frac{\pi }{6}} \right)\) , \(\left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right) – \left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{two}} \correct)} \right)\)
correct substitution A1
e.g. \(\frac{{five\pi }}{vi} – \sin \frac{{5\pi }}{three} – \left[ {\frac{\pi }{6} – \sin \frac{\pi }{3}} \right]\) , \(\left( {\frac{{five\pi }}{half-dozen} – \left( { – \frac{{\sqrt iii }}{two}} \correct)} \correct) – \left( {\frac{\pi }{half-dozen} – \frac{{\sqrt 3 }}{2}} \right)\)
distance is \(\frac{{2\pi }}{iii} + \sqrt 3 \) A1A1 N3
Note: Award A1 for \(\frac{{2\pi }}{3}\) , A1 for \(\sqrt 3 \) .
[five marks]
d.
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Source: https://www.iitianacademy.com/ib-dp-maths-topic-6-6-kinematic-problems-involving-displacement-s-velocity-v-and-acceleration-a-sl-paper-1/
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