June 2025 Paper 1 Q6
6. A scientist is monitoring the flight of sea birds after they leave their nests on a cliff.
The height above the sea, \(h\) metres, of one of the birds is modelled by the equation
\[h = A - B\,t^{1.5} \qquad\qquad h \geqslant 0 \qquad 0 \leqslant t \leqslant T\]where \(t\) seconds is the time after the bird leaves its nest and \(A\), \(B\) and \(T\) are positive constants.
Given that
- the bird was 17.6 m above the sea exactly 4 seconds after leaving its nest
- the bird was 11.9 m above the sea exactly 9 seconds after leaving its nest
Find, according to the model,
| Scheme | Marks | AO |
|---|---|---|
| Attempts to use \(h = A - B\,t^{1.5}\) to form one correct equation with either \(17.6 = A - B \times 4^{1.5}\) or \(11.9 = A - B \times 9^{1.5}\) | M1 | 3.1b |
| Correct equations \(\begin{aligned}A - 8B &= 17.6\\ A - 27B &= 11.9\end{aligned}\) | A1 | 1.1b |
| Solves simultaneously to find values for \(A\) and \(B\) | dM1 | 1.1b |
| \(h = 20 - 0.3\,t^{1.5}\) | A1 | 3.3 |
| (4) |
Notes
M1: Forms one correct equation either \(17.6 = A - 8B\) or \(11.9 = A - 27B\)
May be unsimplified e.g. \(17.6 = A - B \times 4^{1.5}\) or \(11.9 = A - B \times 9^{1.5}\)
A1: \(17.6 = A - 8B\) and \(11.9 = A - 27B\) which may be unsimplified
dM1: Solves simultaneously to find values for \(A\) and \(B\). It is dependent on the previous method mark so at least one equation must be correct. Do not be concerned with the process as calculators may be used. Score if values for \(A\) and \(B\) are reached from a pair of simultaneous equations.
A1: \(h = 20 - 0.3\,t^{1.5}\) o.e. e.g. \(t = \left(\dfrac{10}{3}(20 - h)\right)^{\frac{2}{3}}\) isw once a correct equation is found.
Requires the complete equation including \(h = \ldots\)
Just stating the values for \(A\) and \(B\) is A0 but allow A1 to be scored if the correct equation is seen in (b) or (c)
| Scheme | Marks | AO |
|---|---|---|
| 20 metres | B1ft | 3.2a |
| (1) |
Notes
B1ft: 20 metres but ft on their \(A\) metres (provided \(A \gt 0\)). Condone if they have a value for \(A\) which is given to a greater degree of accuracy than 3sf but they round this to 3sf or better in (b).
Requires units as well (allow m for metres)
| Scheme | Marks | AO |
|---|---|---|
| \(0 = 20 - 0.3\,T^{1.5}\ \left(\Rightarrow T^{1.5} = \dfrac{200}{3}\right) \Rightarrow T = \ldots\) | M1 | 3.4 |
| \((T =)\ 16.4\) | A1 | 1.1b |
| (2) | ||
| (7 marks) |
Notes
Do not be concerned with the use of t or T. If awrt 16.4 is seen in (a) it must be seen or used in (c) to score.
M1: Sets \(\text{``}20\text{''} - \text{``}0.3\text{''}\,T^{1.5} = 0\) and proceeds to a value or expression for \(T\) e.g. \(\left(\dfrac{\text{``}20\text{''}}{\text{``}0.3\text{''}}\right)^{\frac{2}{3}}\)
Do not be concerned by the use of any inequalities instead of “=”
You do not need to be concerned by the mechanics of the rearrangement, they just need to achieve a value or expression.
If no equation is seen then may be implied by a correct value for \(T\) (to the nearest integer e.g. awrt “16”) or a correct expression for \(T\) for their \(A\) and \(B\). You may need to check \(\left(\dfrac{\text{``}20\text{''}}{\text{``}0.3\text{''}}\right)^{\frac{2}{3}}\) on your calculator. They may also achieve this by trial and improvement.
A1: \((T =)\) awrt 16.4 including \(\left(\dfrac{200}{3}\right)^{\frac{2}{3}}\) or exact equivalent e.g. \(\sqrt[\frac{3}{2}]{\dfrac{200}{3}}\) ignore any units for time. If an exact value is given condone the radical symbol not fully covering the fraction provided it is not clearly e.g. \(\dfrac{\sqrt[\frac{3}{2}]{200}}{3}\)
Condone \((0 \leqslant)\ T \leqslant\) awrt 16.4 o.e. but not \((0 \leqslant)\ T \lt\) awrt 16.4
Do not accept any greater than (or equal) inequalities e.g. \(T \geqslant\) awrt 16.4
isw if there is a written response to (c) once a correct value or valid expression for \(T\) is seen.