| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Applied quadratic optimization |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question using quadratic functions. Part (a) requires simple interpretation, (b) is solving a quadratic equation, (c) is completing the square (a standard AS technique), and (d) reads off the maximum from vertex form. All parts are routine applications of basic techniques with no problem-solving insight required. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02z Models in context: use functions in modelling |
| Answer | Marks | Guidance |
|---|---|---|
| The tower is 7 m tall, the stone is at height 7m at time 0 | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 4.60\) seconds | M1, M1, A1 | Recognising that the values of \(t\) such that \(h(t) = 0\) need to be found; Uses the quadratic formula to find the values of \(t\); States that \(t = 4.60\) sec is the correct answer OR crosses out the other answer OR explains that \(t\) must be positive |
| Answer | Marks | Guidance |
|---|---|---|
| \(h(t) = 7 + 21t - 4.9t^2 = -4.9(t - \frac{15}{7})^2 + 29.5\) with \(B = 4.9\) and \(C = \frac{15}{7}\) | B1, M1, A1 | Achieves \(7 + 21t - 4.9t^2 = -4.9(t \pm k)^2 \pm \cdots\) or states that \(B = 4.9\); Deals correctly with first two terms of \(7 + 21t - 4.9t^2\); All coefficients correct |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum height: 29.5 m, Time: \(t = \frac{15}{7}\) seconds | B1, B1 | (2 marks) |
### Part a:
The tower is 7 m tall, the stone is at height 7m at time 0 | B1 | (1 mark)
### Part b:
$t = 4.60$ seconds | M1, M1, A1 | Recognising that the values of $t$ such that $h(t) = 0$ need to be found; Uses the quadratic formula to find the values of $t$; States that $t = 4.60$ sec is the correct answer OR crosses out the other answer OR explains that $t$ must be positive
### Part c:
$h(t) = 7 + 21t - 4.9t^2 = -4.9(t - \frac{15}{7})^2 + 29.5$ with $B = 4.9$ and $C = \frac{15}{7}$ | B1, M1, A1 | Achieves $7 + 21t - 4.9t^2 = -4.9(t \pm k)^2 \pm \cdots$ or states that $B = 4.9$; Deals correctly with first two terms of $7 + 21t - 4.9t^2$; All coefficients correct
### Part d:
Maximum height: 29.5 m, Time: $t = \frac{15}{7}$ seconds | B1, B1 | (2 marks)\includegraphics{figure_1}
A stone is thrown over level ground from the top of a tower, $X$.
The height, $h$, in meters, of the stone above the ground level after $t$ seconds is modelled by the function.
$$h(t) = 7 + 21t - 4.9t^2, \quad t \geq 0$$
A sketch of $h$ against $t$ is shown in Figure 1.
Using the model,
\begin{enumerate}[label=(\alph*)]
\item give a physical interpretation of the meaning of the constant term 7 in the model. [1]
\item find the time taken after the stone is thrown for it to reach ground level. [3]
\item Rearrange $h(t)$ into the form $A - B(t - C)^2$, where $A$, $B$ and $C$ are constants to be found. [3]
\item Using your answer to part c or otherwise, find the maximum height of the stone above the ground, and the time after which this maximum height is reached. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q6 [9]}}