Higher January 2022 Paper 2 Q24
24 A particle \(P\) moves along a straight line that passes through the fixed point \(O\)
The displacement, \(x\) metres, of \(P\) from \(O\) at time \(t\) seconds, where \(t \geqslant 0\), is given by
\(x = 4t^3 - 27t + 8\)
The direction of motion of \(P\) reverses when \(P\) is at the point \(A\) on the line.
The acceleration of \(P\) at the instant when \(P\) is at \(A\) is \(a\) m/s2
Find the value of \(a\)
(5)
| Scheme | Marks |
|---|---|
| \((v =)\;12t^2 - 27\;(= 0)\) | M1 |
\(t^2 = \dfrac{27}{12}\;\left(= \dfrac{9}{4}\right)\) oe or \((3)(2t + 3)(2t - 3)\;(= 0)\) | M1 |
| \(\sqrt{\dfrac{9}{4}}\) oe \(\left(= \dfrac{3}{2}\right)\) or \(\pm\sqrt{\dfrac{9}{4}}\) oe \(\left(= \pm\dfrac{3}{2}\right)\) | A1 |
| \((a =)\;24t\) | M1 |
| 36 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: Correct differentiation
M1: dep M1 first stage to solve \(v = 0\) by rearranging, factorising, quadratic formula, or completing the square
A1: Correct value of \(t\) (allow \(\pm\))
M1: dep 1st M1 for differentiating \(v\)
A1: correct answer