| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a standard M2 kinematics question requiring integration of acceleration to find velocity, solving a quadratic equation, then integrating velocity to find displacement. While it involves multiple steps and careful calculation (including finding when the particle changes direction), the techniques are routine for M2 students with no novel problem-solving required. Slightly above average difficulty due to the multi-part nature and need to consider motion in both directions. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(v = \int a\,dt = 3t^2 - 10t + c\) | M1 A1 |
| When \(t=0\), \(v=3\) so \(c=3\) \(\therefore v = 3t^2 - 10t + 3\) | M1 A1 | |
| \(v = 0\) when \((3t-1)(t-3) = 0\) \(\therefore t = \frac{1}{3}, 3\) | M1 A1 | |
| (b) | \(s = \int v\,dt = t^3 - 5t^2 + 3t + k\) | M1 A1 |
| When \(t=0\), \(s=0\) so \(k=0\) \(\therefore s = t^3 - 5t^2 + 3t\) | A1 | |
| Disp. when \(t=\frac{4}{3}\) is \((\frac{1}{3})^3 - 5(\frac{1}{3})^2 + 3(\frac{1}{3}) = \frac{13}{27}\) | M1 A1 | |
| Disp. when \(t=2\) is \((2)^3 - 5(2)^2 + 3(2) = -6\) | A1 | |
| Dist. travelled \(= 2 \times \frac{13}{27} + 6 = 6\frac{26}{27}\text{ m}\) | A1 | (13 marks total) |
(a) | $v = \int a\,dt = 3t^2 - 10t + c$ | M1 A1 |
| When $t=0$, $v=3$ so $c=3$ $\therefore v = 3t^2 - 10t + 3$ | M1 A1 |
| $v = 0$ when $(3t-1)(t-3) = 0$ $\therefore t = \frac{1}{3}, 3$ | M1 A1 |
(b) | $s = \int v\,dt = t^3 - 5t^2 + 3t + k$ | M1 A1 |
| When $t=0$, $s=0$ so $k=0$ $\therefore s = t^3 - 5t^2 + 3t$ | A1 |
| Disp. when $t=\frac{4}{3}$ is $(\frac{1}{3})^3 - 5(\frac{1}{3})^2 + 3(\frac{1}{3}) = \frac{13}{27}$ | M1 A1 |
| Disp. when $t=2$ is $(2)^3 - 5(2)^2 + 3(2) = -6$ | A1 |
| Dist. travelled $= 2 \times \frac{13}{27} + 6 = 6\frac{26}{27}\text{ m}$ | A1 | (13 marks total)
---5. A particle $P$ moves in a straight line with an acceleration of $( 6 t - 10 ) \mathrm { m } \mathrm { s } ^ { - 2 }$ at time $t$ seconds. Initially $P$ is at $O$, a fixed point on the line, and has velocity $3 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $t$ for which the velocity of $P$ is zero.
\item Show that, during the first two seconds, $P$ travels a distance of $6 \frac { 26 } { 27 } \mathrm {~m}$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [13]}}