Edexcel M1 2010 June — Question 5 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2010
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeTwo-particle meeting or overtaking
DifficultyStandard +0.3 This is a standard M1 kinematics problem involving two vehicles with different motion profiles meeting at the same point. While it requires careful setup of equations using suvat and area under speed-time graphs, the problem-solving approach is routine for M1 students. The 12 marks reflect length rather than conceptual difficulty—it's slightly easier than average A-level questions overall due to being methodical rather than requiring insight.
Spec3.02a 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

Two cars \(P\) and \(Q\) are moving in the same direction along the same straight horizontal road. Car \(P\) is moving with constant speed 25 m s\(^{-1}\). At time \(t = 0\), \(P\) overtakes \(Q\) which is moving with constant speed 20 m s\(^{-1}\). From \(t = T\) seconds, \(P\) decelerates uniformly, coming to rest at a point \(X\) which is 800 m from the point where \(P\) overtook \(Q\). From \(t = 25\) s, \(Q\) decelerates uniformly, coming to rest at the same point \(X\) at the same instant as \(P\).
  1. Sketch, on the same axes, the speed-time graphs of the two cars for the period from \(t = 0\) to the time when they both come to rest at the point \(X\). [4]
  2. Find the value of \(T\). [8]

Part (a)
AnswerMarks
Shape (both)B1
CrossB1
Meet on \(t\)-axisB1
Figures 25, 20, T, 25B1
[4]
Part (b)
AnswerMarks
For \(Q\): \(20\left(\frac{t + 25}{2}\right) = 800\)M1 A1
\(t = 55\)DM1 A1
For \(P\): \(25\left(\frac{T + 55}{2}\right) = 800\)M1 A1
solving for \(T\); \(T = 9\)DM1 A1
[8]
[12]
## Part (a)

Shape (both) | B1 |
Cross | B1 |
Meet on $t$-axis | B1 |
Figures 25, 20, T, 25 | B1 |
| [4] |

## Part (b)

For $Q$: $20\left(\frac{t + 25}{2}\right) = 800$ | M1 A1 |
$t = 55$ | DM1 A1 |

For $P$: $25\left(\frac{T + 55}{2}\right) = 800$ | M1 A1 |
solving for $T$; $T = 9$ | DM1 A1 |
| [8] |
| [12] |
Two cars $P$ and $Q$ are moving in the same direction along the same straight horizontal road. Car $P$ is moving with constant speed 25 m s$^{-1}$. At time $t = 0$, $P$ overtakes $Q$ which is moving with constant speed 20 m s$^{-1}$. From $t = T$ seconds, $P$ decelerates uniformly, coming to rest at a point $X$ which is 800 m from the point where $P$ overtook $Q$. From $t = 25$ s, $Q$ decelerates uniformly, coming to rest at the same point $X$ at the same instant as $P$.

\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same axes, the speed-time graphs of the two cars for the period from $t = 0$ to the time when they both come to rest at the point $X$.
[4]

\item Find the value of $T$.
[8]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2010 Q5 [12]}}
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