| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | January |
| Marks | 6 |
| 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 question requiring students to find when velocity is zero (factorized form given), integrate to find displacement for each interval, and sum absolute values. The factorization is already done and the method is routine textbook material, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = (2t-3)(t-2) = 0\) | M1 | Solve for \(v = 0\) |
| \(t = \frac{3}{2}\) or \(2\) | A1 | Both values |
| \(\int 2t^2 - 7t + 6\, dt\) | M1 | Use of \(s = \int v\, dt\) |
| \(= \frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\ (+C)\) | A1 | Correct integration |
| \(s = \int_0^{\frac{3}{2}} v\, dt - \int_{\frac{3}{2}}^{2} v\, dt + \int_{2}^{3} v\, dt\) | M1 | Correct strategy for distance. Accept equivalent e.g. \(s = \int_0^3 v\, dt + 2\left |
| \(= \left[\frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\right]_0^{\frac{3}{2}} - \left[\frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\right]_{\frac{3}{2}}^{2} + \left[\frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\right]_2^3\) | \(= \frac{27}{8} + \frac{1}{24} + \frac{7}{6}\) | |
| \(= \frac{55}{12}\) | A1 | 4.6 or better from correct working |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = (2t-3)(t-2) = 0$ | M1 | Solve for $v = 0$ |
| $t = \frac{3}{2}$ or $2$ | A1 | Both values |
| $\int 2t^2 - 7t + 6\, dt$ | M1 | Use of $s = \int v\, dt$ |
| $= \frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\ (+C)$ | A1 | Correct integration |
| $s = \int_0^{\frac{3}{2}} v\, dt - \int_{\frac{3}{2}}^{2} v\, dt + \int_{2}^{3} v\, dt$ | M1 | Correct strategy for distance. Accept equivalent e.g. $s = \int_0^3 v\, dt + 2\left|\int_{\frac{3}{2}}^2 v\, dt\right|$ |
| $= \left[\frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\right]_0^{\frac{3}{2}} - \left[\frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\right]_{\frac{3}{2}}^{2} + \left[\frac{2}{3}t^3 - \frac{7}{2}t^2 + 6t\right]_2^3$ | | $= \frac{27}{8} + \frac{1}{24} + \frac{7}{6}$ |
| $= \frac{55}{12}$ | A1 | 4.6 or better from correct working |
---\begin{enumerate}
\item A particle $P$ moves along a straight line. At time $t = 0 , P$ passes the point $A$ on the line and at time $t$ seconds the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ where
\end{enumerate}
$$v = ( 2 t - 3 ) ( t - 2 )$$
At $t = 3 , P$ reaches the point $B$. Find the total distance moved by $P$ as it travels from $A$ to $B$.\\
(6)\\
\hfill \mbox{\textit{Edexcel M2 2017 Q3 [6]}}