| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Standard +0.3 This is a straightforward non-constant acceleration problem requiring differentiation to find when acceleration is zero (part a) and integration to find displacement when velocity is zero (part b). Both parts use standard AS-level calculus techniques with no conceptual challenges beyond routine application. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(v = 3t - 2t^2 + 14\) and differentiate | M1 | Multiply out and attempt to differentiate, at least one power decreasing |
| \(a = \frac{dv}{dt} = 3 - 4t\) | A1 | Correct expression |
| \(3 - 4t = 0\) and solve for \(t\) | M1 | Equate their \(a\) to 0 and solve for \(t\) |
| \(t = \frac{3}{4}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Solve \(v = 0\) to find \(t = \frac{7}{2}\) | M1 | Uses \(v = 0\) to obtain a value of \(t\) |
| \(v = 3t - 2t^2 + 14\) and integrate | M1 | Attempt to integrate, at least one power increasing |
| \(s = \frac{3t^2}{2} - \frac{2t^3}{3} + 14t\) | A1 | Correct expression |
| Substitute \(t = \frac{7}{2}\) into \(s\) expression | M1 | Must have come from using \(v=0\); M0 if using suvat |
| \(s = \frac{931}{24} = 38\frac{19}{24} = 38.79166...\) (m) | A1 | Accept 39 or better |
# Question 3:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $v = 3t - 2t^2 + 14$ and differentiate | M1 | Multiply out and attempt to differentiate, at least one power decreasing |
| $a = \frac{dv}{dt} = 3 - 4t$ | A1 | Correct expression |
| $3 - 4t = 0$ and solve for $t$ | M1 | Equate their $a$ to 0 and solve for $t$ |
| $t = \frac{3}{4}$ | A1 | cao |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| Solve $v = 0$ to find $t = \frac{7}{2}$ | M1 | Uses $v = 0$ to obtain a value of $t$ |
| $v = 3t - 2t^2 + 14$ and integrate | M1 | Attempt to integrate, at least one power increasing |
| $s = \frac{3t^2}{2} - \frac{2t^3}{3} + 14t$ | A1 | Correct expression |
| Substitute $t = \frac{7}{2}$ into $s$ expression | M1 | Must have come from using $v=0$; M0 if using suvat |
| $s = \frac{931}{24} = 38\frac{19}{24} = 38.79166...$ (m) | A1 | Accept 39 or better |\begin{enumerate}
\item A particle $P$ moves along a straight line such that at time $t$ seconds, $t \geqslant 0$, after leaving the point $O$ on the line, the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of $P$ is modelled as
\end{enumerate}
$$v = ( 7 - 2 t ) ( t + 2 )$$
(a) Find the value of $t$ at the instant when $P$ stops accelerating.\\
(b) Find the distance of $P$ from $O$ at the instant when $P$ changes its direction of motion.
In this question, solutions relying on calculator technology are not acceptable.
\hfill \mbox{\textit{Edexcel AS Paper 2 2020 Q3 [9]}}