CAIE M1 2017 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2017
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeConstant acceleration with algebraic unknowns
DifficultyStandard +0.3 This is a standard kinematics problem requiring application of SUVAT equations with two constraints (equal distances AB=BC, different times). The 'show that' format guides students to the answer, and the method is straightforward once the correct equations are set up. Slightly above average due to the algebraic manipulation needed, but well within typical M1 scope.
Spec3.02d Constant acceleration: SUVAT formulae

A car travels along a straight road with constant acceleration. It passes through points \(A\), \(B\) and \(C\). The car passes point \(A\) with velocity 14 m s\(^{-1}\). The two sections \(AB\) and \(BC\) are of equal length. The times taken to travel along \(AB\) and \(BC\) are 5 s and 3 s respectively.
  1. Write down an expression for the distance \(AB\) in terms of the acceleration of the car. Write down a similar expression for the distance \(AC\). Hence show that the acceleration of the car is 4 m s\(^{-2}\). [4]
  2. Find the speed of the car as it passes point \(C\). [2]

Question 3:

AnswerMarks Guidance
3(i)s = 14 × 5 + ½a × 52
ABB1 or s = ½(14 + 14 + 5a) × 5 OE
AB
s = 14 × 8 + ½a × 82
AnswerMarks Guidance
ACB1 or s = ½(14 + 14 + 8a) × 8 OE
AC
AnswerMarks Guidance
[112 + 32a = 2(70 + 12.5a)]M1 Using AC = 2AB and solving for a or for
substituting a = 4 and finding AB and AC
AnswerMarks Guidance
a = 4 m s–2A1 AG, If substituting a = 4 must show
AB = 120 and AC = 240 OE
4

AnswerMarks Guidance
3(ii)[v = 14 + 4 × 8] M1
or any complete method to find v
AnswerMarks
Velocity = 46 m s-1A1
2
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
--- 3(i) ---
3(i) | s = 14 × 5 + ½a × 52
AB | B1 | or s = ½(14 + 14 + 5a) × 5 OE
AB
s = 14 × 8 + ½a × 82
AC | B1 | or s = ½(14 + 14 + 8a) × 8 OE
AC
[112 + 32a = 2(70 + 12.5a)] | M1 | Using AC = 2AB and solving for a or for
substituting a = 4 and finding AB and AC
a = 4 m s–2 | A1 | AG, If substituting a = 4 must show
AB = 120 and AC = 240 OE
4
--- 3(ii) ---
3(ii) | [v = 14 + 4 × 8] | M1 | Use of v = u + at
or any complete method to find v
Velocity = 46 m s-1 | A1
2
Question | Answer | Marks | Guidance
A car travels along a straight road with constant acceleration. It passes through points $A$, $B$ and $C$. The car passes point $A$ with velocity 14 m s$^{-1}$. The two sections $AB$ and $BC$ are of equal length. The times taken to travel along $AB$ and $BC$ are 5 s and 3 s respectively.

\begin{enumerate}[label=(\roman*)]
\item Write down an expression for the distance $AB$ in terms of the acceleration of the car. Write down a similar expression for the distance $AC$. Hence show that the acceleration of the car is 4 m s$^{-2}$. [4]
\item Find the speed of the car as it passes point $C$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2017 Q3 [6]}}