| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a standard M2 variable acceleration question requiring two integrations with given initial conditions. The proportionality constant must be found using the velocity condition at t=3, then a second integration yields displacement. While it involves multiple steps, the techniques are routine for M2 students and the algebra is straightforward, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a \propto (3t^2 - 5)\) \(\therefore a = k(3t^2 - 5)\) | M1 | |
| \(v = \int a\,dt = k(t^3 - 5t) + c\) | M1 A1 | |
| When \(t = 0\), \(v = 0\) so \(c = 0\) | A1 | |
| When \(t = 3\), \(v = 3\) so \(3 = k(27 - 15)\) \(\therefore k = \frac{1}{4}\) | M1 A1 | |
| \(a = \frac{1}{4}(3t^2 - 5)\) | A1 | |
| (b) \(s = \int v\,dt = \frac{1}{4}(\frac{1}{4}t^4 - \frac{5}{2}t^2) + c\) | M1 A1 | |
| When \(t = 0\), \(s = 0\) so \(c = 0\) \(\therefore s = \frac{1}{4}(\frac{1}{4}t^4 - \frac{5}{2}t^2)\) | M1 | |
| \(s = \frac{1}{16}t^2(t^2 - 10) = \frac{1}{16}t^2(t^2 - 10)\) | A1 | (11 marks) |
**(a)** $a \propto (3t^2 - 5)$ $\therefore a = k(3t^2 - 5)$ | M1 |
$v = \int a\,dt = k(t^3 - 5t) + c$ | M1 A1 |
When $t = 0$, $v = 0$ so $c = 0$ | A1 |
When $t = 3$, $v = 3$ so $3 = k(27 - 15)$ $\therefore k = \frac{1}{4}$ | M1 A1 |
$a = \frac{1}{4}(3t^2 - 5)$ | A1 |
**(b)** $s = \int v\,dt = \frac{1}{4}(\frac{1}{4}t^4 - \frac{5}{2}t^2) + c$ | M1 A1 |
When $t = 0$, $s = 0$ so $c = 0$ $\therefore s = \frac{1}{4}(\frac{1}{4}t^4 - \frac{5}{2}t^2)$ | M1 |
$s = \frac{1}{16}t^2(t^2 - 10) = \frac{1}{16}t^2(t^2 - 10)$ | A1 | (11 marks)4. A particle $P$ moves in a straight horizontal line such that its acceleration at time $t$ seconds is proportional to $\left( 3 t ^ { 2 } - 5 \right)$.
Given that at time $t = 0 , P$ is at rest at the origin $O$ and that at time $t = 3$, its velocity is $3 \mathrm {~ms} ^ { - 1 }$,
\begin{enumerate}[label=(\alph*)]
\item find, in $\mathrm { m } \mathrm { s } ^ { - 2 }$, the acceleration of $P$ in terms of $t$,
\item show that the displacement of the particle, $s$ metres, from $O$ at time $t$ is given by
$$s = \frac { 1 } { 16 } t ^ { 2 } \left( t ^ { 2 } - 10 \right)$$
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [11]}}