Higher January 2021 Paper 1R Q20
20 A particle \(P\) is moving along a straight line.
The fixed point \(O\) lies on the line.
At time \(t\) seconds \((t \geqslant 0)\), the displacement of \(P\) from \(O\) is \(s\) metres where
\(s = t^3 - 9t^2 + 33t - 6\)
Find the minimum speed of \(P\).
(5)
| Scheme | Marks |
|---|---|
| \((v =)\; 3t^2 - 9 \times 2t + 33\) | M1 |
\((a =)\; 3 \times 2t - \text{‘}18\text{’}\) or \((t =)\; -\dfrac{-18}{2 \times 3}\left(= \dfrac{18}{6}\right)\) or \((v =)\; 3\left[(t - 3)^2 - (3)^2\right](+33)\) or \((v =)\; 3\left[(t - 3)^2 - (3)^2\;(+11)\right]\) | M1 |
\(6t - 18 = 0\) or \(t = 3\) or \((v =)\; 3\left[(t - 3)^2 - (3)^2\right] + 33\) or \((v =)\; 3\left[(t - 3)^2 - (3)^2 + 11\right]\) | M1 |
\(3 \times \text{‘}3\text{’}^2 - 18 \times \text{‘}3\text{’} + 33\) or \((v =)\; 3(t - 3)^2 + 6\) or \((v =)\; 3\left[(t - 3)^2 + 2\right]\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 6 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for differentiating at least 2 terms correctly
M1: dep ft must be a two term linear equation
or
for the use of \((t =)\; -\dfrac{b}{2a}\)
or
for a correct first step for completing the square on at least a two term quadratic
M1: dep on at least M2 for equating their acceleration to 0
or
for a correct method for completing the square on at least a two term quadratic
M1: dep on at least M2 for substituting their \(t\) into \(v\)
or
for a seeing a correct simplified expression after completing the square