| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic differentiation for acceleration, solving a quadratic equation using a given condition, and integration for distance. All steps are routine AS-level techniques with no novel problem-solving required, making it easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differentiate \(v\) w.r.t. \(t\) | M1 | Differentiate, with both powers decreasing by 1 |
| \(a = \frac{dv}{dt} = 10 - 2t\) isw | A1 | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solve problem using \(v = 0\) when \(t = 6\) | M1 | Put \(t = 6\) OR use \((t-6)(t-x) = t^2 - 10t + k\) oe |
| \(0 = 10t - t^2 - 24\) | A1 | Correct expression (unsimplified) for \(v\) OR \(v = (t-6)(t-4)\) |
| Solve quadratic oe to find other value of \(t\) | M1 | Put \(v = 0\) to give quadratic in \(t\) and solve for other value of \(t\) |
| \(t = 4\) | A1 | \(t = 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integrate \(v\) or \(-v\) w.r.t. \(t\) | M1 | Integrate, with at least two powers increasing by 1 (allow if only two terms integrated) |
| \(5t^2 - \frac{1}{3}t^3 - 24t\) | A1 | Correct expression |
| Total distance \(= -\left[5t^2 - \frac{1}{3}t^3 - 24t\right]_0^4 + \left[5t^2 - \frac{1}{3}t^3 - 24t\right]_4^6\) | M1 | Complete method to find total distance |
| \(\frac{116}{3}\) (m) | A1 | Accept 39(m) or better |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiate $v$ w.r.t. $t$ | M1 | Differentiate, with both powers decreasing by 1 |
| $a = \frac{dv}{dt} = 10 - 2t$ isw | A1 | Correct expression |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solve problem using $v = 0$ when $t = 6$ | M1 | Put $t = 6$ **OR** use $(t-6)(t-x) = t^2 - 10t + k$ oe |
| $0 = 10t - t^2 - 24$ | A1 | Correct expression (unsimplified) for $v$ **OR** $v = (t-6)(t-4)$ |
| Solve quadratic oe to find other value of $t$ | M1 | Put $v = 0$ to give quadratic in $t$ and solve for other value of $t$ |
| $t = 4$ | A1 | $t = 4$ |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integrate $v$ or $-v$ w.r.t. $t$ | M1 | Integrate, with at least two powers increasing by 1 (allow if only two terms integrated) |
| $5t^2 - \frac{1}{3}t^3 - 24t$ | A1 | Correct expression |
| Total distance $= -\left[5t^2 - \frac{1}{3}t^3 - 24t\right]_0^4 + \left[5t^2 - \frac{1}{3}t^3 - 24t\right]_4^6$ | M1 | Complete method to find total distance |
| $\frac{116}{3}$ (m) | A1 | Accept 39(m) or better |\begin{enumerate}
\item A particle $P$ moves along a straight line.
\end{enumerate}
At time $t$ seconds, the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of $P$ is modelled as
$$v = 10 t - t ^ { 2 } - k \quad t \geqslant 0$$
where $k$ is a constant.\\
(a) Find the acceleration of $P$ at time $t$ seconds.
The particle $P$ is instantaneously at rest when $t = 6$\\
(b) Find the other value of $t$ when $P$ is instantaneously at rest.\\
(c) Find the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 6$
\hfill \mbox{\textit{Edexcel AS Paper 2 2021 Q2 [10]}}