| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Standard +0.3 This is a standard M1 SUVAT problem requiring a speed-time graph sketch, forming equations from three motion stages, and solving simultaneous equations. While it involves algebraic manipulation with two unknowns (V and T), the approach is methodical and follows a well-practiced template that M1 students routinely encounter. The problem-solving is structured rather than requiring novel insight, making it slightly easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Trapezium shape starting at origin, finishing on \(t\)-axis | B1 | B1 for trapezium shape, starting at origin and finishing on the \(t\)-axis |
| Figures \((V, T, 180)\) correctly positioned | B1 | B1 for \(V\), \(T\) with delineators or marked on top of trapezium or on axes, and 180 correctly positioned |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Time accelerating \(= V/1 = V\) | M1 | M1 for both time accelerating \(= V/1 = V\) and time decelerating \(= V/0.5 = 2V\); M0 if no working shown or if \(V - 0.5\) used |
| Time decelerating \(= V/0.5 = 2V\) | ||
| Time at constant speed \(T = 180 - (2V + V)\) | ||
| \(T = 180 - 3V\) | A1 | A1 for printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{2}(180 + 180 - 3V)V = 4800\) | M1 A1 A1 | M1 for area under graph = 4800 with \(\frac{1}{2}\) present; M0 if suvat formula used |
| \(V^2 - 120V + 3200 = 0\) | A1 | Third A1 for correct quadratic \(= 0\) |
| \((V-40)(V-80) = 0\) | DM1 | DM1 dependent on first M1 for solving quadratic |
| \(V = 40\) or \(80\) or both; since \((180 - 3 \times 80) < 0\), reject \(V = 80\) | A1, M1 | Fourth A1 for \(V = 40\) or \(V = 80\) or both; third M1 for correct reason rejecting \(V = 80\) (only if both values found); allow "Since \(T > 0\), \(V = 40\)" |
# Question 7:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Trapezium shape starting at origin, finishing on $t$-axis | B1 | B1 for trapezium shape, starting at origin and finishing on the $t$-axis |
| Figures $(V, T, 180)$ correctly positioned | B1 | B1 for $V$, $T$ with delineators or marked on top of trapezium or on axes, and 180 correctly positioned |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Time accelerating $= V/1 = V$ | M1 | M1 for both time accelerating $= V/1 = V$ and time decelerating $= V/0.5 = 2V$; M0 if no working shown or if $V - 0.5$ used |
| Time decelerating $= V/0.5 = 2V$ | | |
| Time at constant speed $T = 180 - (2V + V)$ | | |
| $T = 180 - 3V$ | A1 | A1 for printed answer |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}(180 + 180 - 3V)V = 4800$ | M1 A1 A1 | M1 for area under graph = 4800 with $\frac{1}{2}$ present; M0 if suvat formula used |
| $V^2 - 120V + 3200 = 0$ | A1 | Third A1 for correct quadratic $= 0$ |
| $(V-40)(V-80) = 0$ | DM1 | DM1 dependent on first M1 for solving quadratic |
| $V = 40$ or $80$ or both; since $(180 - 3 \times 80) < 0$, reject $V = 80$ | A1, M1 | Fourth A1 for $V = 40$ or $V = 80$ or both; third M1 for correct reason rejecting $V = 80$ (only if both values found); allow "Since $T > 0$, $V = 40$" |
---7. A train moves on a straight horizontal track between two stations $A$ and $B$. The train starts from rest at $A$ and moves with constant acceleration $1 \mathrm {~ms} ^ { - 2 }$ until it reaches a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The train maintains this speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for the next $T$ seconds before slowing down with constant deceleration $0.5 \mathrm {~ms} ^ { - 2 }$, coming to rest at $B$. The journey from $A$ to $B$ takes 180 s and the distance between the stations is 4800 m .
\begin{enumerate}[label=(\alph*)]
\item Sketch a speed-time graph for the motion of the train from $A$ to $B$.
\item Show that $T = 180 - 3 V$.
\item Find the value of $V$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2016 Q7 [11]}}