Higher January 2020 Paper 1R Q23
23 A particle moves along a straight line.
The fixed point \(O\) lies on this line.
The displacement of the particle from \(O\) at time \(t\) seconds, \(t \geqslant 0\), is \(s\) metres where
\[s = t^3 + 4t^2 - 5t + 7\]
At time \(T\) seconds the velocity of \(P\) is \(V\) m/s where \(\;V \geqslant -5\)
Find an expression for \(T\) in terms of \(V\).
Give your expression in the form \(\;\dfrac{-4 + \sqrt{k + mV}}{3}\;\) where \(k\) and \(m\) are integers to be found.
(6)
| Scheme | Marks |
|---|---|
| (\(v\) =) \(3t^2 + 2 \times 4t - 5\) | M1 |
| \(3T^2 + 8T - 5 = V\) OR \(3T^2 + 8T - 5 - V = 0\) | A1 |
\(3\left(T^2 + \dfrac{8}{3}T\right) - 5\) OR \(3\left(T^2 + \dfrac{8}{3}T - \dfrac{5}{3}\right)\) or (\(T\) =) \(\dfrac{-8 \pm \sqrt{8^2 - 4 \times 3 \times (-5 - V)}}{2 \times 3}\) | M1 |
\(\left(T + \dfrac{4}{3}\right)^2 = \left(\dfrac{4}{3}\right)^2 + \dfrac{V + 5}{3}\) or (\(T\) =) \(\dfrac{-8 \pm \sqrt{124 + 12V}}{6}\) | M1 |
\(T = \dfrac{-4}{3} \pm \dfrac{1}{3}\sqrt{16 + 3V + 15}\) or (\(T\) =) \(\dfrac{-8 \pm 2\sqrt{31 + 3V}}{6}\) | M1 |
| \(\dfrac{-4 + \sqrt{31 + 3V}}{3}\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: 2 out of 3 terms differentiated correctly
A1: correct equation
M1: attempt to complete the square OR use quadratic formula (condone one sign error in \(a\), \(b\) or \(c\) and ft their quadratic with mistake in \(a\) or \(b\))
(condone + instead of ±)
M1: sight of this method mark implies the previous M1
(condone + instead of ±)
(ft their quadratic with mistake in \(a\) or \(b\))
M1: (condone + instead of ±)
(ft their quadratic with mistake in \(a\) or \(b\))
A1: accept \(k = 31\) and \(m = 3\)