| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (algebraic) |
| Difficulty | Standard +0.3 This is a standard M1 two-stage motion problem requiring SUVAT equations and equating distances. Part (b) is straightforward equation solving, and part (c) guides students to a specific quadratic form. The multi-step nature and two-phase acceleration add slight complexity, but this is a typical textbook exercise with no novel insight required—slightly easier than average for A-level. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct graph shape: horizontal line, two lines with positive gradient (second less steep than first), both ending at same \(t\)-value | B1 | No solid vertical line at end; intermediate solid vertical lines allowed |
| Values 40, 15, \(15+T\) correctly placed | B1 | Allow appropriate delineators |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(40 = 4t_1 \Rightarrow t_1 = 10\) | M1 A1 | Complete method to give equation in \(t_1\) only |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(60 \ (\text{m s}^{-1})\) | B1 | |
| \(60 + T \ (\text{m s}^{-1})\) | B1 ft | \(60 + T\) seen or implied; ft on graph. If \(s = ut + \frac{1}{2}at^2\) used, \(60+T\) not needed |
| \(\frac{1}{2}\times15\times60 + \frac{1}{2}T(60+60+T) = 40(15+T)\) | M1 A2 | Equating distances, equation in \(T\) only; correct structure. M0 if \(\frac{1}{2}\) omitted or a section omitted |
| OR \(\frac{1}{2}\times15\times60 + 60T + \frac{1}{2}T\times T = 40(15+T)\) | ||
| OR \(\frac{1}{2}(T+T+15)\times60 + \frac{1}{2}T\times T = 40(15+T)\) | ||
| \(T^2 + 40T - 300 = 0 \ (k=40)\) | A1 | Correct quadratic with \(k=40\) |
| Total | (6) | NB: If \(T\) taken as end of time period (not \(15+T\)), max: (a) B1B0 (b) M1A1 (c) B1B1ft M1A0A0A0, with \(T\) replaced by \((T-15)\) |
# Question 5:
## Part 5(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct graph shape: horizontal line, two lines with positive gradient (second less steep than first), both ending at same $t$-value | B1 | No solid vertical line at end; intermediate solid vertical lines allowed |
| Values 40, 15, $15+T$ correctly placed | B1 | Allow appropriate delineators |
| **Total** | **(2)** | |
## Part 5(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $40 = 4t_1 \Rightarrow t_1 = 10$ | M1 A1 | Complete method to give equation in $t_1$ only |
| **Total** | **(2)** | |
## Part 5(c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $60 \ (\text{m s}^{-1})$ | B1 | |
| $60 + T \ (\text{m s}^{-1})$ | B1 ft | $60 + T$ seen or implied; ft on graph. If $s = ut + \frac{1}{2}at^2$ used, $60+T$ not needed |
| $\frac{1}{2}\times15\times60 + \frac{1}{2}T(60+60+T) = 40(15+T)$ | M1 A2 | Equating distances, equation in $T$ only; correct structure. M0 if $\frac{1}{2}$ omitted or a section omitted |
| **OR** $\frac{1}{2}\times15\times60 + 60T + \frac{1}{2}T\times T = 40(15+T)$ | | |
| **OR** $\frac{1}{2}(T+T+15)\times60 + \frac{1}{2}T\times T = 40(15+T)$ | | |
| $T^2 + 40T - 300 = 0 \ (k=40)$ | A1 | Correct quadratic with $k=40$ |
| **Total** | **(6)** | **NB:** If $T$ taken as end of time period (not $15+T$), max: (a) B1B0 (b) M1A1 (c) B1B1ft M1A0A0A0, with $T$ replaced by $(T-15)$ |
---5. A car travels at a constant speed of $40 \mathrm {~ms} ^ { - 1 }$ in a straight line along a horizontal racetrack. At time $t = 0$, the car passes a motorcyclist who is at rest. The motorcyclist immediately sets off to catch up with the car.
The motorcyclist accelerates at $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 15 s and then accelerates at $1 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for a further $T$ seconds until he catches up with the car.
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same axes, the speed-time graph for the motion of the car and the speed-time graph for the motion of the motorcyclist, from time $t = 0$ to the instant when the motorcyclist catches up with the car.
At the instant when $t = t _ { 1 }$ seconds, the car and the motorcyclist are moving at the same speed.
\item Find the value of $t _ { 1 }$
\item Show that $T ^ { 2 } + k T - 300 = 0$, where $k$ is a constant to be found.
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel M1 2020 Q5 [10]}}