Edexcel M1 2017 June — Question 5 13 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2017
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeSketch velocity-time graph
DifficultyStandard +0.3 This is a standard M1 two-particle kinematics problem requiring sketching speed-time graphs, finding when speeds are equal (simple algebra), and using areas under graphs to find when Q overtakes P. All techniques are routine for M1 students with no novel insight 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

  1. Two trains, \(P\) and \(Q\), move on horizontal parallel straight tracks. Initially both are at rest in a station and level with each other. At time \(t = 0 , P\) starts off and moves with constant acceleration for 10 s up to a speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and then moves at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). At time \(t = 20\), where \(t\) is measured in seconds, train \(Q\) starts to move in the same direction as \(P\). Train \(Q\) accelerates with the same initial constant acceleration as \(P\), up to a speed of \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and then moves at a constant speed of \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Train \(Q\) overtakes \(P\) at time \(t = T\), after both trains have reached their constant speeds.
    1. Sketch, on the same axes, the speed-time graphs of both trains for \(0 \leqslant t \leqslant T\).
    2. Find the value of \(t\) at the instant when both trains are moving at the same speed.
    3. Find the value of \(T\).

Question 5(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
One graph correct shapeB1
Both graphs correct shape, on same sketch and intersecting (with different start times)B1 Figs 10, 20, 25, 40 shown (with 20 as the second start time)
Correct figures shownB1 Ignore all vertical lines
Question 5(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(20 + 10\)M1 Complete method
\(= 30\)A1
Question 5(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\dfrac{40}{t_1 - 20} = \dfrac{25}{10}\)M1 Complete method to find time when \(Q\) reaches \(40\text{ m s}^{-1}\)
Correct unsimplified equationA1
\(\Rightarrow t_1 = 36\)A1
Time to reach \(40\text{ m s}^{-1}\) is \(\dfrac{40}{2.5}(=16)\) (M1A1)
Time from start \(= \dfrac{40}{2.5} + 20 = 36\) (A1) (seen or implied)
\(\dfrac{(T+T-10)}{2}\times25\)M1 Find distance travelled by either train at \(t=T\)
One correctA1
\(\dfrac{(T-20+T-36)}{2}\times40\)A1ft Both correct. Follow their 36
Equate and solve for \(T\)dM1
\(T = 66\tfrac{1}{3}\)A1 Accept 66 or better
## Question 5(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| One graph correct shape | B1 | |
| Both graphs correct shape, on same sketch and intersecting (with different start times) | B1 | Figs 10, 20, 25, 40 shown (with 20 as the second start time) |
| Correct figures shown | B1 | Ignore all vertical lines |

## Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $20 + 10$ | M1 | Complete method |
| $= 30$ | A1 | |

## Question 5(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{40}{t_1 - 20} = \dfrac{25}{10}$ | M1 | Complete method to find time when $Q$ reaches $40\text{ m s}^{-1}$ |
| Correct unsimplified equation | A1 | |
| $\Rightarrow t_1 = 36$ | A1 | |
| Time to reach $40\text{ m s}^{-1}$ is $\dfrac{40}{2.5}(=16)$ (M1A1) | | |
| Time from start $= \dfrac{40}{2.5} + 20 = 36$ (A1) | | (seen or implied) |
| $\dfrac{(T+T-10)}{2}\times25$ | M1 | Find distance travelled by either train at $t=T$ |
| One correct | A1 | |
| $\dfrac{(T-20+T-36)}{2}\times40$ | A1ft | Both correct. Follow their 36 |
| Equate and solve for $T$ | dM1 | |
| $T = 66\tfrac{1}{3}$ | A1 | Accept 66 or better |
\begin{enumerate}
  \item Two trains, $P$ and $Q$, move on horizontal parallel straight tracks. Initially both are at rest in a station and level with each other. At time $t = 0 , P$ starts off and moves with constant acceleration for 10 s up to a speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and then moves at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t = 20$, where $t$ is measured in seconds, train $Q$ starts to move in the same direction as $P$. Train $Q$ accelerates with the same initial constant acceleration as $P$, up to a speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and then moves at a constant speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Train $Q$ overtakes $P$ at time $t = T$, after both trains have reached their constant speeds.\\
(a) Sketch, on the same axes, the speed-time graphs of both trains for $0 \leqslant t \leqslant T$.\\
(b) Find the value of $t$ at the instant when both trains are moving at the same speed.\\
(c) Find the value of $T$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2017 Q5 [13]}}