Edexcel M1 2014 June — Question 3 13 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeMulti-phase journey: find unknown speed or time
DifficultyModerate -0.3 This is a standard M1 SUVAT question with multiple stages requiring systematic application of kinematic equations. While it involves several phases and algebraic manipulation with an unknown V, each step follows routine procedures: using s=½(u+v)t to find V from the given distance, then applying SUVAT formulae to each subsequent stage. The multi-stage nature adds length but not conceptual difficulty—it's slightly easier than average due to being a textbook-style structured problem with clear signposting.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

3. A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in 20 seconds. It moves at constant speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) for the next 30 seconds, then moves with constant deceleration \(\frac { 1 } { 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\) until it has speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It moves at speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) for the next 15 seconds and then moves with constant deceleration \(\frac { 1 } { 3 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\) until it comes to rest.
  1. Sketch, in the space below, a speed-time graph for this journey. In the first 20 seconds of this journey the car travels 140 m . Find
  2. the value of \(V\),
  3. the total time for this journey,
  4. the total distance travelled by the car.

Question 3:
Part (a)
AnswerMarks Guidance
WorkingMarks Guidance
Shape of graph for \(0 \leq t \leq 50\)B1
Shape of graph for \(t > 50\)B1
Values \(V, 8, 15, 20, 30\) appropriately usedB1
Part (b)
AnswerMarks Guidance
WorkingMarks Guidance
\(140 = \frac{1}{2} \times 20 \times V\)M1 Use of area under graph (must have '\(\frac{1}{2}\)') or suvat to obtain equation in \(V\) only
\(V = 14\)A1
Part (c)
AnswerMarks Guidance
WorkingMarks Guidance
\(8 = V - \frac{1}{3}t_1\) and/or \(0 = 8 - \frac{1}{3}t_2\)M1 First M1 for use of either equation
\(t_1 = 12\), (and/or \(t_2 = 24\))A1 First A1 for either value
Total time \(= 20 + 30 + t_1 + 15 + t_2 = 101\) (seconds)DM1 A1 Second M1 dependent on first M1; must include all 5 time intervals
Part (d)
AnswerMarks Guidance
WorkingMarks Guidance
Total distance \(= 140 + 30V + \frac{V+8}{2}t_1 + 15\times 8 + \frac{1}{2}\times 8 \times t_2\)M1A2ft First M1 for expression including all parts of motion; where triangle or trapezium used a '\(\frac{1}{2}\)' must be seen. A2 ft on \(V\), \(t_1\), \(t_2\) (−1 each error)
\(= 140 + 30\times14 + 11\times12 + 15\times8 + 24\times4\)
\(= 908\ \text{(m)}\)A1
# Question 3:

## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Shape of graph for $0 \leq t \leq 50$ | B1 | |
| Shape of graph for $t > 50$ | B1 | |
| Values $V, 8, 15, 20, 30$ appropriately used | B1 | |

## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $140 = \frac{1}{2} \times 20 \times V$ | M1 | Use of area under graph (must have '$\frac{1}{2}$') or suvat to obtain equation in $V$ only |
| $V = 14$ | A1 | |

## Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $8 = V - \frac{1}{3}t_1$ and/or $0 = 8 - \frac{1}{3}t_2$ | M1 | First M1 for use of either equation |
| $t_1 = 12$, (and/or $t_2 = 24$) | A1 | First A1 for either value |
| Total time $= 20 + 30 + t_1 + 15 + t_2 = 101$ (seconds) | DM1 A1 | Second M1 dependent on first M1; must include all 5 time intervals |

## Part (d)
| Working | Marks | Guidance |
|---------|-------|----------|
| Total distance $= 140 + 30V + \frac{V+8}{2}t_1 + 15\times 8 + \frac{1}{2}\times 8 \times t_2$ | M1A2ft | First M1 for expression including all parts of motion; where triangle or trapezium used a '$\frac{1}{2}$' must be seen. A2 ft on $V$, $t_1$, $t_2$ (−1 each error) |
| $= 140 + 30\times14 + 11\times12 + 15\times8 + 24\times4$ | | |
| $= 908\ \text{(m)}$ | A1 | |

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3. A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 20 seconds. It moves at constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for the next 30 seconds, then moves with constant deceleration $\frac { 1 } { 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ until it has speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It moves at speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for the next 15 seconds and then moves with constant deceleration $\frac { 1 } { 3 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ until it comes to rest.
\begin{enumerate}[label=(\alph*)]
\item Sketch, in the space below, a speed-time graph for this journey.

In the first 20 seconds of this journey the car travels 140 m .

Find
\item the value of $V$,
\item the total time for this journey,
\item the total distance travelled by the car.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2014 Q3 [13]}}