Higher June 2022 Paper 1 Q17
17 A particle \(P\) moves along a straight line.
The fixed point \(O\) lies on this line.
The displacement of \(P\) from \(O\) at time \(t\) seconds, \(t \geqslant 1\), is \(s\) metres where
\[s = 4t^2 + \dfrac{125}{t}\]The velocity of \(P\) at time \(t\) seconds, \(t \geqslant 1\), is \(v\) m / s
Work out the distance of \(P\) from \(O\) at the instant when \(v = 0\)
(5)
| Scheme | Marks |
|---|---|
| \(8t\) or \(\pm 125t^{-2}\) oe | M1 |
| \(8t - 125t^{-2}\) oe or \(8t - \dfrac{125}{t^2}\) oe | A1 |
| \(8t - 125t^{-2} = 0\) and \((t =)\;\sqrt[3]{\dfrac{125}{8}}\;(= 2.5)\) | M1 |
| \(4\left(\text{``}{2.5}\text{''}\right)^2 + \dfrac{125}{\text{``}{2.5}\text{''}}\) | M1 |
| 75 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for differentiating one term correctly
A1: for both terms correct
M1: for equating their \(8t \pm at^{-2}\) oe or \(bt \pm 125t^{-2}\) oe to zero and solving for \(t\) ie must have correct powers of \(t\) and at least one correct coefficient and correct isolation of \(t\)
M1: dep on previous M mark for substituting into \(s\)