Higher June 2018 Paper 2 Q16
16 \(R\) is proportional to \(t^2\)
The graph shows the relationship between \(R\) and \(t\) for \(0 \leqslant t \leqslant 4\)
(a) Find a formula for \(R\) in terms of \(t\). (3)
Given also that \(R = \dfrac{8}{5x}\)
(b) show that \(t\) is inversely proportional to \(\sqrt{x}\) for \(t \gt 0\) (2)
| Scheme | Marks |
|---|---|
| \(R = kt^2\) oe | M1 |
| eg \(10 = k \times 2^2\) or \(40 = k \times 4^2\) or \(k = 2\tfrac{1}{2}\) | M1 |
Working required Answer: \(R = \dfrac{5}{2}t^2\) | A1 |
| (3) |
Notes
M1: Equation consistent with \(R \propto t^2\)
M1: Substitute values at any point on the graph or find the value of \(k\). (Implies first M1.) Allow readings from graph for \(t\) ± 0.1 and \(R\) ± 1
A1: Award for \(R = kt^2\) if the value of \(k\) is shown clearly in (a) or (b).
| Scheme | Marks |
|---|---|
| \(\dfrac{8}{5x} = \text{``}{\dfrac{5}{2}t^2}\text{''}\) | M1 |
Working required Answer: \(t = \dfrac{0.8}{\sqrt{x}}\) | A1 |
| (2) | |
| (5 marks) |
Notes
M1: ft dep on answer of the form \(R = kt^2\)
A1: ft dep on answer of the form \(R = kt^2\)
Simplification of constant is not required. eg accept \(t = \sqrt{\dfrac{16}{25}} \times \dfrac{1}{\sqrt{x}}\)
[allow other clear arguments that clearly shows \(t\) is inversely proportional to \(\sqrt{x}\)]