Edexcel M1 2003 November — Question 4 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2003
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with algebraic unknowns
DifficultyModerate -0.3 This is a standard M1 kinematics question involving speed-time graphs and constant acceleration equations. Part (a) is routine sketching, part (b) uses area under graph equals distance, and parts (c)-(d) apply SUVAT equations with careful attention to timing. While multi-step, all techniques are textbook standard with no novel problem-solving required, making it slightly easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

A car starts from rest at a point \(S\) on a straight racetrack. The car moves with constant acceleration for 20 s, reaching a speed of 25 m s\(^{-1}\). The car then travels at a constant speed of 25 m s\(^{-1}\) for 120 s. Finally it moves with constant deceleration, coming to rest at a point \(F\).
  1. In the space below, sketch a speed-time graph to illustrate the motion of the car. [2]
The distance between \(S\) and \(F\) is 4 km.
  1. Calculate the total time the car takes to travel from \(S\) to \(F\). [3]
A motorcycle starts at \(S\), 10 s after the car has left \(S\). The motorcycle moves with constant acceleration from rest and passes the car at a point \(P\) which is 1.5 km from \(S\). When the motorcycle passes the car, the motorcycle is still accelerating and the car is moving at a constant speed. Calculate
  1. the time the motorcycle takes to travel from \(S\) to \(P\), [5]
  2. the speed of the motorcycle at \(P\). [2]

(a)
AnswerMarks Guidance
ShapeB1 (2)
FigsB1
(b)
AnswerMarks Guidance
\(\frac{1}{2}(T + 120) \times 25 = 4000\)M1 A1
or \(\left[\frac{1}{2} \cdot 20 \cdot 25 + 120 \cdot 25 + \frac{1}{2}(T - 140) \cdot 25 = 4000\right]\) A1
\(\rightarrow T = 200\) s (3)
(c)
AnswerMarks Guidance
Car: \(\frac{1}{2} \cdot 20 \cdot 25 + 25(t - 20) = 1500\)M1 A1, A1
\(\rightarrow t = 70\) sM1 A1
Hence motorcycle travels for 60 s (5)
(d)
AnswerMarks Guidance
\(1500 = \left(\frac{0 + v}{2}\right) \cdot 60\)M1
\(v = 50 \text{ m s}^{-1}\) A1
Total: 12 marks
## (a)
Shape | B1 | (2)
Figs | B1 |

## (b)
$\frac{1}{2}(T + 120) \times 25 = 4000$ | M1 | A1
or $\left[\frac{1}{2} \cdot 20 \cdot 25 + 120 \cdot 25 + \frac{1}{2}(T - 140) \cdot 25 = 4000\right]$ | | A1
$\rightarrow T = 200$ s | | (3)

## (c)
Car: $\frac{1}{2} \cdot 20 \cdot 25 + 25(t - 20) = 1500$ | M1 | A1, A1
$\rightarrow t = 70$ s | M1 | A1
Hence motorcycle travels for 60 s | | (5)

## (d)
$1500 = \left(\frac{0 + v}{2}\right) \cdot 60$ | M1 |
$v = 50 \text{ m s}^{-1}$ | | A1

**Total: 12 marks**

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A car starts from rest at a point $S$ on a straight racetrack. The car moves with constant acceleration for 20 s, reaching a speed of 25 m s$^{-1}$. The car then travels at a constant speed of 25 m s$^{-1}$ for 120 s. Finally it moves with constant deceleration, coming to rest at a point $F$.

\begin{enumerate}[label=(\alph*)]
\item In the space below, sketch a speed-time graph to illustrate the motion of the car. [2]
\end{enumerate}

The distance between $S$ and $F$ is 4 km.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Calculate the total time the car takes to travel from $S$ to $F$. [3]
\end{enumerate}

A motorcycle starts at $S$, 10 s after the car has left $S$. The motorcycle moves with constant acceleration from rest and passes the car at a point $P$ which is 1.5 km from $S$. When the motorcycle passes the car, the motorcycle is still accelerating and the car is moving at a constant speed. Calculate

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item the time the motorcycle takes to travel from $S$ to $P$, [5]
\item the speed of the motorcycle at $P$. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2003 Q4 [12]}}