| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a straightforward M2 kinematics question requiring differentiation of a polynomial to find velocity and acceleration, solving a quadratic equation, and calculating distance travelled. All techniques are standard and mechanical with no novel problem-solving required, making it slightly easier than average but still requiring multiple steps and careful attention to distance vs displacement. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v = 3t^2 - 14t + 8 = (3t - 2)(t - 4)\) \(\quad v = 0: t = \frac{2}{3}, t = 4\) | M1 A1 M1 A1 | |
| \(P\) is turning round (changing direction) | A1 | |
| (b) \(s(3) = -12, s(4) = -16, s(5) = -10\), so dist \(= 4 + 6 = 10\) m | M1 A1 A1 | |
| (c) \(a = 6t - 14\) \(\quad a = -2\) when \(t = 2\) | M1 A1 | |
| Negative acceleration acting on negative velocity, so speeding up | B1 | Total: 11 marks |
**(a)** $v = 3t^2 - 14t + 8 = (3t - 2)(t - 4)$ $\quad v = 0: t = \frac{2}{3}, t = 4$ | M1 A1 M1 A1 |
$P$ is turning round (changing direction) | A1 |
**(b)** $s(3) = -12, s(4) = -16, s(5) = -10$, so dist $= 4 + 6 = 10$ m | M1 A1 A1 |
**(c)** $a = 6t - 14$ $\quad a = -2$ when $t = 2$ | M1 A1 |
Negative acceleration acting on negative velocity, so speeding up | B1 | Total: 11 marks7. A particle $P$ moves in a straight line so that its displacement $s$ metres from a fixed point $O$ at time $t$ seconds is given by the formula $s = t ^ { 3 } - 7 t ^ { 2 } + 8 t$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $t$ when the velocity of $P$ equals zero, and briefly describe what is happening to $P$ at these times.
\item Find the distance travelled by $P$ between the times $t = 3$ and $t = 5$.
\item Find the value of $t$ when the acceleration of $P$ is $- 2 \mathrm {~ms} ^ { - 2 }$. Briefly explain the significance of a negative acceleration at this time.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [11]}}