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Date	May 2017	Marks available	5	Reference code	17M.2.AHL.TZ1.H_11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Determine	Question number	H_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, $v\,{\text{m}}{{\text{s}}^{ - 1}}$ , $t$ seconds after jumping from the plane, can be modelled by the function

$v(t) = \left\{ {\begin{array}{*{20}{l}} {9.8t{\text{,}}}&{0 \leqslant t \leqslant 10} \\ {\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }},}&{t > 10} \end{array}} \right.$

His velocity when he reaches the ground is $2.8{\text{ m}}{{\text{s}}^{ - 1}}$ .

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

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$v(15) = \frac{{98}}{{\sqrt {1 + {{(15 - 10)}^2}} }}$ (M1)

$v(15) = 19.2{\text{ }}({\text{m}}{{\text{s}}^{ - 1}})$ A1

[2 marks]

a.

$\int\limits_0^{10} {9.8t\,{\text{d}}t}$ (M1)

$= 490{\text{ }}({\text{m}})$ A1

[2 marks]

b.

$\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }} = 2.8$ (M1)

$t = 44.985 \ldots {\text{ }}({\text{s}})$ A1

$h = 490 + \int\limits_{10}^{44.9...} {\frac{{98}}{{\sqrt {1 + {{(t - 10)}^2}} }}{\text{d}}t}$ (M1)(A1)

$h = 906{\text{ (m}})$ A1

[5 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Syllabus sections

Topic 5—Calculus » AHL 5.13—Kinematic problems

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Topic 5—Calculus