| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.hl.TZ2.7 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Determine | Question number | 7 | Adapted from | N/A |
Question
A point P moves in a straight line with velocity \(v\) ms−1 given by \(v\left( t \right) = {{\text{e}}^{ - t}} - 8{t^2}{{\text{e}}^{ - 2t}}\) at time t seconds, where t ≥ 0.
Determine the first time t1 at which P has zero velocity.
[2]
a.
Find an expression for the acceleration of P at time t.
[2]
b.i.
Find the value of the acceleration of P at time t1.
[1]
b.ii.
Markscheme
attempt to solve \(v\left( t \right) = 0\) for t or equivalent (M1)
t1 = 0.441(s) A1
[2 marks]
a.
\(a\left( t \right) = \frac{{{\text{d}}v}}{{{\text{d}}t}} = - {{\text{e}}^{ - t}} - 16t{{\text{e}}^{ - 2t}} + 16{t^2}{{\text{e}}^{ - 2t}}\) M1A1
Note: Award M1 for attempting to differentiate using the product rule.
[2 marks]
b.i.
\(a\left( {{t_1}} \right) = - 2.28\) (ms−2) A1
[1 mark]
b.ii.
Examiners report
[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
Syllabus sections
Topic 6 - Core: Calculus » 6.6 » Kinematic problems involving displacement \(s\), velocity \(v\) and acceleration \(a\).
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