| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | January |
| Marks | 8 |
| 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 M3 variable acceleration question requiring integration of acceleration to find velocity (with careful attention to the initial condition sign), then finding when velocity is zero to calculate total distance. While it requires multiple steps and understanding of direction, the integration is straightforward and the method is a textbook exercise for this module. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(a = \frac{dv}{dt} = 6 - 2t\) | ||
| \(v = 6t - t^2 \ (+c)\) | M1A1 | Using \(a = \frac{dv}{dt}\) and integrating; correct integration w/wo constant |
| \(t=0, v=-8 \Rightarrow c=-8\) | Find constant | |
| \(v = 6t - t^2 - 8\) | A1 (3) | Correct statement for velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v = 6t - t^2 - 8 = 0\) | ||
| \((t-2)(t-4) = 0\) | ||
| \(t=2, \ t=4\) | M1 | Find times when \(P\) is at rest |
| \(s = \int(6t - t^2 - 8)dt = 3t^2 - \frac{1}{3}t^3 - 8t \ (+c)\) | M1 | Integrate \(v\) to obtain expression for \(s\) |
| \(\left[3t^2 - \frac{1}{3}t^3 - 8t\right]_2^4 = 48 - \frac{64}{3} - 32 - \left(12 - \frac{8}{3} - 16\right) = 1\frac{1}{3}\) | A1 | Correct displacement for either time interval |
| \(\left[3t^2 - \frac{1}{3}t^3 - 8t\right]_0^2 = 12 - \frac{8}{3} - 16 - (0) = -6\frac{2}{3}\) | A1 | Correct displacement for second time interval |
| Total distance \(= 8\) m | A1ft (5) [8] | Add 2 positive distances |
## Question 2:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $a = \frac{dv}{dt} = 6 - 2t$ | | |
| $v = 6t - t^2 \ (+c)$ | M1A1 | Using $a = \frac{dv}{dt}$ and integrating; correct integration w/wo constant |
| $t=0, v=-8 \Rightarrow c=-8$ | | Find constant |
| $v = 6t - t^2 - 8$ | A1 (3) | Correct statement for velocity |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $v = 6t - t^2 - 8 = 0$ | | |
| $(t-2)(t-4) = 0$ | | |
| $t=2, \ t=4$ | M1 | Find times when $P$ is at rest |
| $s = \int(6t - t^2 - 8)dt = 3t^2 - \frac{1}{3}t^3 - 8t \ (+c)$ | M1 | Integrate $v$ to obtain expression for $s$ |
| $\left[3t^2 - \frac{1}{3}t^3 - 8t\right]_2^4 = 48 - \frac{64}{3} - 32 - \left(12 - \frac{8}{3} - 16\right) = 1\frac{1}{3}$ | A1 | Correct displacement for either time interval |
| $\left[3t^2 - \frac{1}{3}t^3 - 8t\right]_0^2 = 12 - \frac{8}{3} - 16 - (0) = -6\frac{2}{3}$ | A1 | Correct displacement for second time interval |
| Total distance $= 8$ m | A1ft (5) [8] | Add 2 positive distances |
---2. A particle $P$ is moving in a straight line. At time $t$ seconds, the distance of $P$ from a fixed point $O$ on the line is $x$ metres and the acceleration of $P$ is $( 6 - 2 t ) \mathrm { m } \mathrm { s } ^ { - 2 }$ in the direction of $x$ increasing. When $t = 0 , P$ is moving towards $O$ with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $P$ in terms of $t$.
\item Find the total distance travelled by $P$ in the first 4 seconds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q2 [8]}}