Edexcel M1 2002 January — Question 4 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2002
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeMulti-phase journey: find unknown speed or time
DifficultyStandard +0.3 This is a standard M1 SUVAT problem requiring speed-time graph sketching and area calculations. While it involves multiple vehicles and requires careful bookkeeping of the different phases of motion, the mathematical techniques are routine (area under graph = distance) with no conceptual surprises. The multi-part structure and need to set up equations systematically makes it slightly above average for M1, but still a textbook-style question.
Spec3.02a Kinematics language: position, displacement, velocity, acceleration3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

A motor scooter and a van set off along a straight road. They both start from rest at the same time and level with each other. The scooter accelerates with constant acceleration until it reaches its top speed of 20 m s\(^{-1}\). It then maintains a constant speed of 20 m s\(^{-1}\). The van accelerates with constant acceleration for 10 s until it reaches its top speed \(V\) m s\(^{-1}\), \(V > 20\). It then maintains a constant speed of \(V\) m s\(^{-1}\). The van draws level with the scooter when the scooter has been travelling for 40 s at its top speed. The total distance travelled by each vehicle is then 850 m.
  1. Sketch on the same diagram the speed-time graphs of both vehicles to illustrate their motion from the time when they start to the time when the van overtakes the scooter. [3]
  2. Find the time for which the scooter is accelerating. [3]
  3. Find the top speed of the van. [3]

Part (b)
Scavenger: \(dv/dt \text{ travelled} = a \text{ under graph}\)
\(850 = \frac{1}{2}T \cdot 20 + 20 \cdot 40\)
AnswerMarks Guidance
\(\Rightarrow T = 55\)M1 A1 A1 (3)
Part (c)
Van: \(850 = \frac{1}{2}V \cdot 10 + V(40-5)\)
AnswerMarks Guidance
\(\Rightarrow V = 21.25 \text{ m s}^{-1}\)M1 A1 M(c) A1 (3) (4)
## Part (b)
Scavenger: $dv/dt \text{ travelled} = a \text{ under graph}$

$850 = \frac{1}{2}T \cdot 20 + 20 \cdot 40$

$\Rightarrow T = 55$ | M1 A1 | A1 (3)

## Part (c)
Van: $850 = \frac{1}{2}V \cdot 10 + V(40-5)$

$\Rightarrow V = 21.25 \text{ m s}^{-1}$ | M1 A1 M(c) | A1 (3) (4)

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A motor scooter and a van set off along a straight road. They both start from rest at the same time and level with each other. The scooter accelerates with constant acceleration until it reaches its top speed of 20 m s$^{-1}$. It then maintains a constant speed of 20 m s$^{-1}$. The van accelerates with constant acceleration for 10 s until it reaches its top speed $V$ m s$^{-1}$, $V > 20$. It then maintains a constant speed of $V$ m s$^{-1}$. The van draws level with the scooter when the scooter has been travelling for 40 s at its top speed. The total distance travelled by each vehicle is then 850 m.

\begin{enumerate}[label=(\alph*)]
\item Sketch on the same diagram the speed-time graphs of both vehicles to illustrate their motion from the time when they start to the time when the van overtakes the scooter. [3]
\item Find the time for which the scooter is accelerating. [3]
\item Find the top speed of the van. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2002 Q4 [9]}}