CAIE M1 2023 November — Question 7 12 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeSketching velocity-time graphs
DifficultyStandard +0.3 This is a straightforward variable acceleration question requiring standard calculus techniques: differentiate to find acceleration, solve a quadratic for stationary points, substitute to find velocities, sketch a cubic graph, and integrate (accounting for sign changes) to find distance. All steps are routine M1 procedures with no novel insight required, making it slightly easier than average.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration
A particle \(X\) travels in a straight line. The velocity of \(X\) at time \(t\) s after leaving a fixed point \(O\) is denoted by \(v\) m s\(^{-1}\), where $$v = -0.1t^3 + 1.8t^2 - 6t + 5.6.$$ The acceleration of \(X\) is zero at \(t = p\) and \(t = q\), where \(p < q\).
  1. Find the value of \(p\) and the value of \(q\). [4]
It is given that the velocity of \(X\) is zero at \(t = 14\).
  1. Find the velocities of \(X\) at \(t = p\) and at \(t = q\), and hence sketch the velocity-time graph for the motion of \(X\) for \(0 \leq t \leq 15\). [3]
  2. Find the total distance travelled by \(X\) between \(t = 0\) and \(t = 15\). [5]
Question 7:
AnswerMarks Guidance
7(a)Attempt to differentiate v *M1
in coefficient in at least one term
(which must be the same term);
allow unsimplified; allow p or q
for t.
v
a= is M0.
t
 dv 
a= =3−0.1t3−1+21.8t2−1−6t1−1 =−0.3t2 +3.6t−6
AnswerMarks Guidance
 dt A1 May be unsimplified.
Allow p or q fort.
dv
Setting a= =0 and attempt to solve a 3 term quadratic for t.
dt
 dv 
a= =0 3t2 −36t+60=0t2 −12t+20=0
 
AnswerMarks Guidance
 dt DM1 Allow p or q fort.
Must get 2 values or numerical
expressions for t from their three
term quadratic.
If using quadratic formula, must be
the correct formula. If factorising,
when brackets expanded, 2 terms
correct.
AnswerMarks
p=2, q=10A1
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)Velocities are 0m s-1 and 25.6m s-1 B1
Curve with single minimum turning point followed by single maximum turning point*B1 Ignore placement of graph on axes
Not a cusp for the minimum point
or maximum point.
AnswerMarks
DB1All correct in 1st and 4th quadrant.
Must go from convex to concave.
Need to label 2 and 14 on the
t-axis where the curve meets the
t-axis.
Do not need to show exact
velocities at t = 0 or
t = 10 or t = 15.
Ignore graph outside 0 t 15.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(c)Attempt to integrate v *M1
in coefficient in at least one term
(which must be the same term);
s=vt is M0.
0.1 1.8 6
(s=)− t3+1+ t2+1− t1+1+5.6t(+c)
4 3 2
AnswerMarks Guidance
=−0.025t4 +0.6t3−3t2 +5.6t [+c]A1 May be unsimplified.
Attempt distance from t=0 to t=14 [= 176.4]DM1 Correct use of limits 0 and 14 for
their s,i.e. F(14)−F(0)
May see limits 0 to 2 and 2 to 14
used but must be
( F(14)−F(2)) ( F(2)−F(0))
+ .
AnswerMarks Guidance
Attempt distance from t=14 to t=15 [= (-) 8.025]DM1 Correct use of limits 14 and 15 for
( F(15)−F(14))
their s,i.e.  .
18
For reference F(2)= =3.6,
5
882
F(14)= =176.4 and
5
1347
F(15)= =168.375.
8
7377

Total distance =1 76.4+8.025=184.425 m =

AnswerMarks Guidance
40A1 www
Condone 184 or better.
AnswerMarks Guidance
QuestionAnswer Marks
7(c)SC for those who show no integration. Max 3 marks.
14
 ( −0.1t3+1.8t2 −6t+5.6 ) dt=176.4
AnswerMarks
0B1
15
 ( −0.1t3+1.8t2 −6t+5.6 ) dt=−8.025
14
15
OR  ( −0.1t3 +1.8t2 −6t+5.6 ) dt =8.025
AnswerMarks
14B1
7377

Total distance =1 76.4+8.025=184.425 m =

AnswerMarks Guidance
40B1 Condone 184 or better.
SC for those who show no integration and don’t consider the 2 areas. Max 1 mark.
15
7377
 −0.1t3+1.8t2 −6t+5.6dt=184.425 m =
40
AnswerMarks Guidance
0B1 Condone 184 or better.
5
Question 7:
--- 7(a) ---
7(a) | Attempt to differentiate v | *M1 | Decrease power by 1 and a change
in coefficient in at least one term
(which must be the same term);
allow unsimplified; allow p or q
for t.
v
a= is M0.
t
 dv 
a= =3−0.1t3−1+21.8t2−1−6t1−1 =−0.3t2 +3.6t−6
 dt  | A1 | May be unsimplified.
Allow p or q fort.
dv
Setting a= =0 and attempt to solve a 3 term quadratic for t.
dt
 dv 
a= =0 3t2 −36t+60=0t2 −12t+20=0
 
 dt  | DM1 | Allow p or q fort.
Must get 2 values or numerical
expressions for t from their three
term quadratic.
If using quadratic formula, must be
the correct formula. If factorising,
when brackets expanded, 2 terms
correct.
p=2, q=10 | A1
4
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Velocities are 0m s-1 and 25.6m s-1 | B1 | SOI
Curve with single minimum turning point followed by single maximum turning point | *B1 | Ignore placement of graph on axes
Not a cusp for the minimum point
or maximum point.
DB1 | All correct in 1st and 4th quadrant.
Must go from convex to concave.
Need to label 2 and 14 on the
t-axis where the curve meets the
t-axis.
Do not need to show exact
velocities at t = 0 or
t = 10 or t = 15.
Ignore graph outside 0 t 15.
3
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | Attempt to integrate v | *M1 | Increase power by 1 and a change
in coefficient in at least one term
(which must be the same term);
s=vt is M0.
0.1 1.8 6
(s=)− t3+1+ t2+1− t1+1+5.6t(+c)
4 3 2
=−0.025t4 +0.6t3−3t2 +5.6t [+c] | A1 | May be unsimplified.
Attempt distance from t=0 to t=14 [= 176.4] | DM1 | Correct use of limits 0 and 14 for
their s,i.e. F(14)−F(0)
May see limits 0 to 2 and 2 to 14
used but must be
( F(14)−F(2)) ( F(2)−F(0))
+ .
Attempt distance from t=14 to t=15 [= (-) 8.025] | DM1 | Correct use of limits 14 and 15 for
( F(15)−F(14))
their s,i.e.  .
18
For reference F(2)= =3.6,
5
882
F(14)= =176.4 and
5
1347
F(15)= =168.375.
8
7377
Total distance =1 76.4+8.025=184.425 m =
40 | A1 | www
Condone 184 or better.
Question | Answer | Marks | Guidance
7(c) | SC for those who show no integration. Max 3 marks.
14
 ( −0.1t3+1.8t2 −6t+5.6 ) dt=176.4
0 | B1
15
 ( −0.1t3+1.8t2 −6t+5.6 ) dt=−8.025
14
15
OR  ( −0.1t3 +1.8t2 −6t+5.6 ) dt =8.025
14 | B1
7377
Total distance =1 76.4+8.025=184.425 m =
40 | B1 | Condone 184 or better.
SC for those who show no integration and don’t consider the 2 areas. Max 1 mark.
15
7377
 −0.1t3+1.8t2 −6t+5.6dt=184.425 m =
40
0 | B1 | Condone 184 or better.
5
A particle $X$ travels in a straight line. The velocity of $X$ at time $t$ s after leaving a fixed point $O$ is denoted by $v$ m s$^{-1}$, where
$$v = -0.1t^3 + 1.8t^2 - 6t + 5.6.$$

The acceleration of $X$ is zero at $t = p$ and $t = q$, where $p < q$.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$ and the value of $q$. [4]
\end{enumerate}

It is given that the velocity of $X$ is zero at $t = 14$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the velocities of $X$ at $t = p$ and at $t = q$, and hence sketch the velocity-time graph for the motion of $X$ for $0 \leq t \leq 15$. [3]

\item Find the total distance travelled by $X$ between $t = 0$ and $t = 15$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q7 [12]}}
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