| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2018 |
| Session | June |
| 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 mechanics question requiring differentiation to find velocity, solving v=0 for rest times, calculating distance with consideration of direction changes, and analyzing the sign of displacement. While it has multiple parts and requires careful handling of the distance calculation, all techniques are routine for AS-level mechanics with no novel problem-solving required. |
| Spec | 1.07a Derivative as gradient: of tangent to curve3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Multiply out and differentiate wrt time (or use product rule; must have two terms with correct structure) | M1 | Must have 3 terms and at least two powers going down by 1 |
| \(v = 2t^3 - 3t^2 + t\) | A1 | Correct expression |
| \(2t^3 - 3t^2 + t = 0\) and solve: \(t(2t-1)(t-1)=0\) | DM1 | Dependent on first M1; equating to zero and attempting to solve cubic |
| \(t=0\) or \(t=\frac{1}{2}\) or \(t=1\); any two | A1 | Two of three correct values (two correct answers can imply correct method) |
| All three | A1 | Third value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \(x\) when \(t=0, \frac{1}{2}, 1\) and \(2\): \((0, \frac{1}{32}, 0, 2)\) | M1 | For attempting to find values of \(x\) at their \(t\) values from (a) or at \(t=2\); or integrate \(v\) and sub in at least two \(t\) values |
| Distance \(= \frac{1}{32}+\frac{1}{32}+2\) | M1 | Using correct strategy to combine distances (must have at least 3 distances) |
| \(2\frac{1}{16}\) (m) oe or 2.06 or better | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \frac{1}{2}t^2(t-1)^2\) | M1 | |
| \(\frac{1}{2}\) × perfect square so \(x \geq 0\) i.e. never negative | A1 cso |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Multiply out and differentiate wrt time (or use product rule; must have two terms with correct structure) | M1 | Must have 3 terms and at least two powers going down by 1 |
| $v = 2t^3 - 3t^2 + t$ | A1 | Correct expression |
| $2t^3 - 3t^2 + t = 0$ and solve: $t(2t-1)(t-1)=0$ | DM1 | Dependent on first M1; equating to zero and attempting to solve cubic |
| $t=0$ or $t=\frac{1}{2}$ or $t=1$; any two | A1 | Two of three correct values (two correct answers can imply correct method) |
| All three | A1 | Third value |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $x$ when $t=0, \frac{1}{2}, 1$ and $2$: $(0, \frac{1}{32}, 0, 2)$ | M1 | For attempting to find values of $x$ at their $t$ values from (a) or at $t=2$; or integrate $v$ and sub in at least two $t$ values |
| Distance $= \frac{1}{32}+\frac{1}{32}+2$ | M1 | Using correct strategy to combine distances (must have at least 3 distances) |
| $2\frac{1}{16}$ (m) oe or 2.06 or better | A1 | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{1}{2}t^2(t-1)^2$ | M1 | |
| $\frac{1}{2}$ × perfect square so $x \geq 0$ i.e. never negative | A1 cso | |\begin{enumerate}
\item A particle, $P$, moves along the $x$-axis. At time $t$ seconds, $t \geqslant 0$, the displacement, $x$ metres, of $P$ from the origin $O$, is given by $x = \frac { 1 } { 2 } t ^ { 2 } \left( t ^ { 2 } - 2 t + 1 \right)$\\
(a) Find the times when $P$ is instantaneously at rest.\\
(b) Find the total distance travelled by $P$ in the time interval $0 \leqslant t \leqslant 2$\\
(c) Show that $P$ will never move along the negative $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2018 Q8 [10]}}