Question
The velocity v ms−1 of an object after t seconds is given by \(v(t) = 15\sqrt t – 3t\) , for \(0 \le t \le 25\) .
On the grid below, sketch the graph of v , clearly indicating the maximum point.
(i) Write down an expression for d .
(ii) Hence, write down the value of d .
Answer/Explanation
Markscheme
A1A1A1 N3
Note: Award A1 for approximately correct shape, A1 for right endpoint at \((25{\text{, }}0)\) and A1 for maximum point in circle.
[3 marks]
(i) recognizing that d is the area under the curve (M1)
e.g. \(\int {v(t)} \)
correct expression in terms of t, with correct limits A2 N3
e.g. \(d = \int_0^9 {(15\sqrt t } – 3t){\rm{d}}t\) , \(d = \int_0^9 v {\rm{d}}t\)
(ii) \(d = 148.5\) (m) (accept 149 to 3 sf) A1 N1
[4 marks]
Question
Let \(f(t) = 2{t^2} + 7\) , where \(t > 0\) . The function v is obtained when the graph of f is transformed by
a stretch by a scale factor of \(\frac{1}{3}\) parallel to the y-axis,
followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 4}
\end{array}} \right)\) .
Find \(v(t)\) , giving your answer in the form \(a{(t – b)^2} + c\) .
A particle moves along a straight line so that its velocity in ms−1 , at time t seconds, is given by v . Find the distance the particle travels between \(t = 5.0\) and \(t = 6.8\) .
Answer/Explanation
Markscheme
applies vertical stretch parallel to the y-axis factor of \(\frac{1}{3}\) (M1)
e.g. multiply by \(\frac{1}{3}\) , \(\frac{1}{3}f(t)\) , \(\frac{1}{3} \times 2\)
applies horizontal shift 2 units to the right (M1)
e.g. \(f(t – 2)\) , \(t – 2\)
applies a vertical shift 4 units down (M1)
e.g. subtracting 4, \(f(t) – 4\) , \(\frac{7}{3} – 4\)
\(v(t) = \frac{2}{3}{(t – 2)^2} – \frac{5}{3}\) A1 N4
[4 marks]
recognizing that distance travelled is area under the curve M1
e.g. \(\int {v,\frac{2}{9}} {(t – 2)^3} – \frac{5}{3}t\) , sketch
distance = 15.576 (accept 15.6) A2 N2
[3 marks]
Question
The velocity of a particle in ms−1 is given by \(v = {{\rm{e}}^{\sin t}} – 1\) , for \(0 \le t \le 5\) .
On the grid below, sketch the graph of \(v\) .
Find the total distance travelled by the particle in the first five seconds.
Write down the positive \(t\)-intercept.
Answer/Explanation
Markscheme
A1A1A1 N3
Note: Award A1 for approximately correct shape crossing x-axis with \(3 < x < 3.5\) .
Only if this A1 is awarded, award the following:
A1 for maximum in circle, A1 for endpoints in circle.
[3 marks]
\(t = \pi \) (exact), \(3.14\) A1 N1
[1 mark]
recognizing distance is area under velocity curve (M1)
eg \(s = \int v \) , shading on diagram, attempt to integrate
valid approach to find the total area (M1)
eg \({\text{area A}} + {\text{area B}}\) , \(\int {v{\rm{d}}t – \int {v{\rm{d}}t} } \) , \(\int_0^{3.14} {v{\rm{d}}t + } \int_{3.14}^5 {v{\rm{d}}t} \) , \(\int {\left| v \right|} \)
correct working with integration and limits (accept \({\rm{d}}x\) or missing \({\rm{d}}t\) ) (A1)
eg \(\int_0^{3.14} {v{\rm{d}}t + } \int_5^{3.14} {v{\rm{d}}t} \) , \(3.067 \ldots + 0.878 \ldots \) , \(\int_0^5 {\left| {{{\rm{e}}^{\sin t}} – 1} \right|} \)
distance \( = 3.95\) (m) A1 N3
[4 marks]
Question
A particle moves along a straight line such that its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\), is given by \(v(t) = 10t{{\text{e}}^{ – 1.7t}}\), for \(t \geqslant 0\).
On the grid below, sketch the graph of \(v\), for \(0 \leqslant t \leqslant 4\).
Find the distance travelled by the particle in the first three seconds.
Find the velocity of the particle when its acceleration is zero.
Answer/Explanation
Markscheme
A1A2 N3
Notes: Award A1 for approximately correct domain \(0 \leqslant t \leqslant 4\).
The shape must be approximately correct, with maximum skewed left. Only if the shape is approximately correct, award A2 for all the following approximately correct features, in circle of tolerance where drawn (accept seeing correct coordinates for the maximum, even if point outside circle):
Maximum point, passes through origin, asymptotic to \(t\)-axis (but must not touch the axis).
If only two of these features are correct, award A1.
[3 marks]
valid approach (including \(0\) and \(3\)) (M1)
eg \(\int_0^3 {10t{{\text{e}}^{ – 1.7t}}{\text{d}}t,{\text{ }}\int_0^3 {f(x)} } \), area from \(0\) to \(3\) (may be shaded in diagram)
\({\text{distance}} = 3.33{\text{ (m)}}\) A1 N2
[2 marks]
recognizing acceleration is derivative of velocity (R1)
eg \(a = \frac{{{\text{d}}v}}{{{\text{d}}t}}\), attempt to find \(\frac{{{\text{d}}v}}{{{\text{d}}t}}\), reference to maximum on the graph of \(v\)
valid approach to find \(v\) when \(a = 0\) (may be seen on graph) (M1)
eg \(\frac{{{\text{d}}v}}{{{\text{d}}t}} = 0,{\text{ }}10{{\text{e}}^{ – 1.7t}} – 17t{{\text{e}}^{ – 1.7t}} = 0,{\text{ }}t = 0.588\)
\({\text{velocity}} = 2.16{\text{ (m}}{{\text{s}}^{ – 1}})\) A1 N3
Note: Award R1M1A0 for \((0.588, 216)\) if velocity is not identified as final answer
[3 marks]
Question
A particle moves in a straight line. Its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\), at time \(t\) seconds, is given by
\[v = {\left( {{t^2} – 4} \right)^3},{\text{ for }}0 \leqslant t \leqslant 3.\]
Find the velocity of the particle when \(t = 1\).
Find the value of \(t\) for which the particle is at rest.
Find the total distance the particle travels during the first three seconds.
Show that the acceleration of the particle is given by \(a = 6t{({t^2} – 4)^2}\).
Find all possible values of \(t\) for which the velocity and acceleration are both positive or both negative.
Answer/Explanation
Markscheme
substituting \(t = 1\) into \(v\) (M1)
eg \(v(1),{\text{ }}{\left( {{1^2} – 4} \right)^3}\)
velocity \( = – 27{\text{ }}\left( {{\text{m}}{{\text{s}}^{ – 1}}} \right)\) A1 N2
[2 marks]
valid reasoning (R1)
eg \(v = 0,{\text{ }}{\left( {{t^2} – 4} \right)^3} = 0\)
correct working (A1)
eg \({t^2} – 4 = 0,{\text{ }}t = \pm 2\), sketch
\(t = 2\) A1 N2
[3 marks]
correct integral expression for distance (A1)
eg \(\int_0^3 {\left| v \right|,{\text{ }}\int {\left| {{{\left( {{t^2} – 4} \right)}^3}} \right|,{\text{ }} – \int_0^2 {v{\text{d}}t + \int_2^3 {v{\text{d}}t} } } } \),
\(\int_0^2 {{{\left( {4 – {t^2}} \right)}^3}{\text{d}}t + \int_2^3 {{{\left( {{t^2} – 4} \right)}^3}{\text{d}}t} }\) (do not accept \(\int_0^3 {v{\text{d}}t} \))
\(86.2571\)
\({\text{distance}} = 86.3{\text{ (m)}}\) A2 N3
[3 marks]
evidence of differentiating velocity (M1)
eg \(v'(t)\)
\(a = 3{\left( {{t^2} – 4} \right)^2}(2t)\) A2
\(a = 6t{\left( {{t^2} – 4} \right)^2}\) AG N0
[3 marks]
METHOD 1
valid approach M1
eg graphs of \(v\) and \(a\)
correct working (A1)
eg areas of same sign indicated on graph
\(2 < t \leqslant 3\) (accept \(t > 2\)) A2 N2
METHOD 2
recognizing that \(a \geqslant 0\) (accept \(a\) is always positive) (seen anywhere) R1
recognizing that \(v\) is positive when \(t > 2\) (seen anywhere) (R1)
\(2 < t \leqslant 3\) (accept \(t > 2\)) A2 N2
[4 marks]
Question
A particle starts from point \(A\) and moves along a straight line. Its velocity, \(v\;{\text{m}}{{\text{s}}^{ – 1}}\), after \(t\) seconds is given by \(v(t) = {{\text{e}}^{\frac{1}{2}\cos t}} – 1\), for \(0 \le t \le 4\). The particle is at rest when \(t = \frac{\pi }{2}\).
The following diagram shows the graph of \(v\).
Find the distance travelled by the particle for \(0 \le t \le\ \frac{\pi }{2}\).
Explain why the particle passes through \(A\) again.
Answer/Explanation
Markscheme
correct substitution of function and/or limits into formula (A1)
(accept absence of d\(t\), but do not accept any errors)
eg\(\;\;\;\)\(\int_0^{\frac{\pi }{2}} {v,{\text{ }}\int {\left| {{{\text{e}}^{\frac{1}{2}\cos t}} – 1} \right|{\text{d}}t,{\text{ }}\int {\left( {{{\text{e}}^{\frac{1}{2}\cos t}} – 1} \right)} } } \)
\(0.613747\)
distance is \(0.614{\text{ }}[0.613,{\text{ }}0.614]{\text{ (m)}}\) A1 N2
[2 marks]
METHOD 1
valid attempt to find the distance travelled between \(t = \frac{\pi }{2}\) and \(t = 4\) (M1)
eg\(\;\;\;\)\(\int_{\frac{\pi }{2}}^4 {\left( {{{\text{e}}^{\frac{1}{2}\cos t}} – 1} \right),{\text{ }}\int_0^4 {\left| {{{\text{e}}^{\frac{1}{2}\cos t}} – 1} \right|{\text{d}}t – 0.614} } \)
distance is \(0.719565\) A1
valid reason, referring to change of direction (may be seen in explanation) R1
valid explanation comparing their distances R1
eg\(\;\;\;\)\(0.719565 > 0.614\), distance moving back is more than distance moving forward
Note: Do not award the final R1 unless the A1 is awarded.
particle passes through \(A\) again AG N0
METHOD 2
valid attempt to find displacement (M1)
eg\(\;\;\;\)\(\int_{\frac{\pi }{2}}^4 {\left( {{{\text{e}}^{\frac{1}{2}\cos t}} – 1} \right),{\text{ }}\int_0^4 {\left( {{{\text{e}}^{\frac{1}{2}\cos t}} – 1} \right)} } \)
correct displacement A1
eg\(\;\;\;\)\(-0.719565,{\text{ }}-0.105817\)
recognizing that displacement from \(0\) to \(\frac{\pi }{2}\) is positive R1
eg\(\;\;\;\)displacement = distance from \(0\) to \(\frac{\pi }{2}\)
valid explanation referring to positive and negative displacement R1
eg\(\;\;\;\)\(0.719565 > 0.614\), overall displacement is negative, since displacement after \(\frac{\pi }{2}\) is negative, then particle gone backwards more than forwards
Note: Do not award the final R1 unless the A1 and the first R1 are awarded.
particle passes through A\(A\) again AG N0
[4 marks]
Note: Special Case.
If all working shown, and candidates seem to have misread the question, using [equation], award marks as follows:
(a) correct substitution of function and/or limits into formula (accept absence of dt, but do not accept any errors) A0MR
eg [equation]
\(2.184544\)
distance is \(2.18\) [\(2.18\), \(2.19\)] (m) A1 N0
(b) METHOD 1
valid attempt to find the distance travelled between [equation] M1
eg [equation]
distance is \(1.709638\) A1
reference to change of direction (may be seen in explanation) R1
reasoning/stating particle passes/does not pass through \(A\) again R0
METHOD 2
valid attempt to find displacement M1
eg [equation]
correct displacement A1
eg \(1.709638,{\rm{ }}3.894182\)
recognising that displacement from [ \(0\) to (pi/2] is positive R0
reasoning/stating particle passes/does not pass through \(A\) again R0
With method 2, there is no valid reasoning about whether the particle passes through \(A\) again or not, so they cannot gain the R marks.
Total [6 marks]