| Date | May 2017 | Marks available | 2 | Reference code | 17M.2.hl.TZ1.11 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 11 | Adapted from | N/A |
Question
Xavier, the parachutist, jumps out of a plane at a height of h metres above the ground. After free falling for 10 seconds his parachute opens. His velocity, vms−1, t seconds after jumping from the plane, can be modelled by the function
v(t)={9.8t,0⩽t⩽1098√1+(t−10)2,t>10
His velocity when he reaches the ground is 2.8 ms−1.
Find his velocity when t=15.
[2]
a.
Calculate the vertical distance Xavier travelled in the first 10 seconds.
[2]
b.
Determine the value of h.
[5]
c.
Markscheme
v(15)=98√1+(15−10)2 (M1)
v(15)=19.2 (ms−1) A1
[2 marks]
a.
10∫09.8tdt (M1)
=490 (m) A1
[2 marks]
b.
98√1+(t−10)2=2.8 (M1)
t=44.985… (s) A1
h=490+44.9...∫1098√1+(t−10)2dt (M1)(A1)
h=906 (m) A1
[5 marks]
c.
Examiners report
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a.
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b.
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c.
