Higher January 2021 Paper 2 Q19
19 A particle \(P\) is moving along a straight line.
The fixed point \(O\) lies on this line.
At time \(t\) seconds where \(t \geqslant 0\), the displacement, \(s\) metres, of \(P\) from \(O\) is given by
\[s = t^3 + 5t^2 - 8t + 10\]Find the displacement of \(P\) from \(O\) when \(P\) is instantaneously at rest.
Give your answer in the form \(\dfrac{a}{b}\) where \(a\) and \(b\) are integers.
(5)
| Scheme | Marks |
|---|---|
| \((v =)\;3t^2 + 10t - 8\) | M1 |
| \(3t^2 + 10t - 8 = 0\) | M1 |
\((3t - 2)(t + 4)\;(= 0)\) \((t =)\;\dfrac{2}{3}\) or \((t =)\;-4\) | M1 |
| \((s =)\left(\dfrac{2}{3}\right)^3 + 5 \times \left(\dfrac{2}{3}\right)^2 - 8 \times \dfrac{2}{3} + 10\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{194}{27}\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: For at least 2 terms differentiated correctly
M1: Their \(v = 0\) dep on M1 could be implied by correct values
M1: dep on M1 for correct values for \(t\) or for \(t = \dfrac{2}{3}\)
or
correct method to solve their 3 term quadratic equation:
If factorising, allow brackets which when expanded give 2 out of 3 terms correct (If using formula or completing the square allow one sign error and some simplification – allow as far as eg \(\dfrac{-10 \pm \sqrt{100 + 96}}{6}\) oe
\(3\left(t + \dfrac{5}{3}\right)^2 - \dfrac{48}{3} = 0\))
M1: For \(\dfrac{2}{3}\) (only) substituted into formula for \(s\) or for selecting the value from this substitution or for an answer of 7.185…
A1: oe but numerator and denominator must be integers.