| Date | May 2018 | Marks available | 1 | Reference code | 18M.2.hl.TZ2.7 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
A point P moves in a straight line with velocity v ms−1 given by v(t)=e−t−8t2e−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(t)=0 for t or equivalent (M1)
t1 = 0.441(s) A1
[2 marks]
a.
a(t)=dvdt=−e−t−16te−2t+16t2e−2t M1A1
Note: Award M1 for attempting to differentiate using the product rule.
[2 marks]
b.i.
a(t1)=−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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