| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | October |
| Marks | 10 |
| 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 variable acceleration question requiring integration of acceleration to find velocity (with initial conditions), then finding when v=0 and integrating again for displacement. While it involves multiple steps and careful handling of signs/limits, it follows a well-practiced procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = t^2 - 3t\ (+c)\) | M1, A1 | Integrate \(a\) to obtain \(v\). Condone missing \(C\) |
| \(t=3,\ v=2 \Rightarrow c=2\) | M1 | Substitute to find \(C\) |
| \(v = t^2 - 3t + 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = (t-2)(t-1)\) | M1 | Set their \(v=0\) and solve for \(t\) |
| \(t=1,\ 2\) | A1 | |
| \(s = \displaystyle\int_1^2 (t^2-3t+2)\,dt\) | M1 | Integrate \(v\) to obtain \(s\) |
| \(= \left[\dfrac{1}{3}t^3 - \dfrac{3}{2}t^2 + 2t\right]_1^2\) | A1ft | Condone if limits not seen. Follow their \(v\). |
| \(= -\dfrac{1}{6}\) m | dM1 | Use their \(t\) values as limits. Dependent on preceding M1. |
| Dist \(= \dfrac{1}{6}\) (m) | A1 | \(0.17,\ 0.167\) or better |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = t^2 - 3t\ (+c)$ | M1, A1 | Integrate $a$ to obtain $v$. Condone missing $C$ |
| $t=3,\ v=2 \Rightarrow c=2$ | M1 | Substitute to find $C$ |
| $v = t^2 - 3t + 2$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = (t-2)(t-1)$ | M1 | Set their $v=0$ and solve for $t$ |
| $t=1,\ 2$ | A1 | |
| $s = \displaystyle\int_1^2 (t^2-3t+2)\,dt$ | M1 | Integrate $v$ to obtain $s$ |
| $= \left[\dfrac{1}{3}t^3 - \dfrac{3}{2}t^2 + 2t\right]_1^2$ | A1ft | Condone if limits not seen. Follow their $v$. |
| $= -\dfrac{1}{6}$ m | dM1 | Use their $t$ values as limits. Dependent on preceding M1. |
| Dist $= \dfrac{1}{6}$ (m) | A1 | $0.17,\ 0.167$ or better |
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\item A particle $P$ moves on the $x$-axis. At time $t$ seconds, $t \geqslant 0$, the acceleration of $P$ is\\
$( 2 t - 3 ) \mathrm { m } \mathrm { s } ^ { - 2 }$ in the positive $x$ direction. At time $t$ seconds, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$ direction. When $t = 3 , v = 2$\\
(a) Find $v$ in terms of $t$.\\
(4)
\end{enumerate}
The particle first comes to instantaneous rest at the point $A$ and then comes to instantaneous rest again at the point $B$.\\
(b) Find the distance $A B$.\\
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\hfill \mbox{\textit{Edexcel M2 2017 Q6 [10]}}