| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Standard +0.3 This is a standard two-particle kinematics problem requiring SUVAT equations and speed-time graph interpretation. Part (a) is straightforward calculation, part (b) is routine sketching, and part (c) requires comparing distances traveled but follows a clear method. Slightly above average due to the multi-part nature and the inequality proof in (c), but all techniques are standard M1 material. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| For Q: \(a = \frac{\Delta v}{\Delta t} = \frac{6-0}{6} = 1\) | M1 |
| \(u = 0, v = 4\), use \(v = u + at\): \(4 = 0 + 1 \cdot t\) i.e. \(t = 4\) seconds | M1 A1 |
| Answer | Marks |
|---|---|
| [Graph showing speed vs time with Q reaching 6 m\(s^{-1}\) and P at 4 m\(s^{-1}\), Q accelerating from O to 4 seconds, then constant] | B3 |
| Answer | Marks | Guidance |
|---|---|---|
| Q will catch P when area under Q graph = area under P graph | M1 | |
| \(\therefore \frac{1}{2}(6)(6) + 6(t - 6) = 4t\) | M1 A1 | |
| i.e. \(18 + 6t - 36 = 4t \therefore 2t = 18 \therefore t = 9\) | M1 A1 | |
| After 9 seconds, P has travelled \(4 \times 9 = 36\) cm, | M1 A1 | |
| \(\therefore\) Q reaches top first if \(x > 36\) | M1 A1 | (11) |
**Part (a)**
For Q: $a = \frac{\Delta v}{\Delta t} = \frac{6-0}{6} = 1$ | M1 |
$u = 0, v = 4$, use $v = u + at$: $4 = 0 + 1 \cdot t$ i.e. $t = 4$ seconds | M1 A1 |
**Part (b)**
[Graph showing speed vs time with Q reaching 6 m$s^{-1}$ and P at 4 m$s^{-1}$, Q accelerating from O to 4 seconds, then constant] | B3 |
**Part (c)**
Q will catch P when area under Q graph = area under P graph | M1 |
$\therefore \frac{1}{2}(6)(6) + 6(t - 6) = 4t$ | M1 A1 |
i.e. $18 + 6t - 36 = 4t \therefore 2t = 18 \therefore t = 9$ | M1 A1 |
After 9 seconds, P has travelled $4 \times 9 = 36$ cm, | M1 A1 |
$\therefore$ Q reaches top first if $x > 36$ | M1 A1 | (11)5. Two flies $P$ and $Q$, are crawling vertically up a wall. At time $t = 0$, the flies are at the same height above the ground, with $P$ crawling at a steady speed of $4 \mathrm { cms } ^ { - 1 }$.\\
$Q$ starts from rest at time $t = 0$ and accelerates uniformly to a speed of $6 \mathrm {~cm} \mathrm {~s} ^ { - 1 }$ in 6 seconds. Fly $Q$ then maintains this speed.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ when the two flies are moving at the same speed.
\item Sketch on the same diagram, speed-time graphs to illustrate the motion of the two flies.
Given that the distance of the two flies from the top of the wall at time $t = 0$ is $x \mathrm {~cm}$ and that $Q$ reaches the top of the wall first,
\item show that $x > 36$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q5 [11]}}