| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| 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 mechanics question requiring differentiation of a velocity function (with fractional powers) to find acceleration, and integration to find displacement. Part (a) is routine calculus; part (b) requires integrating √t and evaluating at limits, which is standard AS-level technique. The 'show that' format and algebraic manipulation adds slight difficulty, but overall this is slightly easier than average for AS mechanics. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate \(2t - 7\sqrt{t} + 6\) with respect to \(t\) | M1 | Both powers of \(t\) decreasing by 1 |
| \(2 - \dfrac{7}{2\sqrt{t}}\) oe | A1 | Any equivalent form |
| When \(t = 4\), \(a = 0.25\ \text{ms}^{-2}\) | A1 | Any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate \(2t - 7\sqrt{t} + 6\) with respect to \(t\) | M1 | At least two powers of \(t\) increasing by 1 |
| \(t^2 - \dfrac{14}{3}t^{\frac{3}{2}} + 6t\ (+C)\) | A1 | Correct integration; accept without \(+C\) and unsimplified |
| Use limits to find \(XY\) | DM1 | Correct use of limits: \(\left(2^2 - \dfrac{14}{3}\times 2^{\frac{3}{2}} + 6\times2\right) - \left(1^2 - \dfrac{14}{3}\times1^{\frac{3}{2}} + 6\times1\right)\); condone missing second bracket; allow subtraction either way and decimals |
| \((XY =)\ \dfrac{1}{3}\left(41 - 28\sqrt{2}\right)\) metres | A1* | Obtain from correct working with at least one more line including a \(\sqrt{2}\) term; not available if decimals used |
# Question 2:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate $2t - 7\sqrt{t} + 6$ with respect to $t$ | M1 | Both powers of $t$ decreasing by 1 |
| $2 - \dfrac{7}{2\sqrt{t}}$ oe | A1 | Any equivalent form |
| When $t = 4$, $a = 0.25\ \text{ms}^{-2}$ | A1 | Any equivalent form |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate $2t - 7\sqrt{t} + 6$ with respect to $t$ | M1 | At least two powers of $t$ increasing by 1 |
| $t^2 - \dfrac{14}{3}t^{\frac{3}{2}} + 6t\ (+C)$ | A1 | Correct integration; accept without $+C$ and unsimplified |
| Use limits to find $XY$ | DM1 | Correct use of limits: $\left(2^2 - \dfrac{14}{3}\times 2^{\frac{3}{2}} + 6\times2\right) - \left(1^2 - \dfrac{14}{3}\times1^{\frac{3}{2}} + 6\times1\right)$; condone missing second bracket; allow subtraction either way and decimals |
| $(XY =)\ \dfrac{1}{3}\left(41 - 28\sqrt{2}\right)$ metres | A1* | Obtain from correct working with at least one more line including a $\sqrt{2}$ term; not available if decimals used |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
A particle is moving along a straight line.\\
At time t seconds, $\mathrm { t } > 0$, the velocity of the particle is $\mathrm { Vms } ^ { - 1 }$, where
$$v = 2 t - 7 \sqrt { t } + 6$$
(a) Find the acceleration of the particle when $t = 4$
When $\mathrm { t } = 1$ the particle is at the point X .\\
When $\mathrm { t } = 2$ the particle is at the point Y .
Given that the particle does not come to instantaneous rest in the interval $1 < \mathrm { t } < 2$\\
(b) show that $X Y = \frac { 1 } { 3 } ( 41 - 28 \sqrt { 2 } )$ metres.
\hfill \mbox{\textit{Edexcel AS Paper 2 2024 Q2 [7]}}