AS 91579 · 2024 paper · integrals & areas

Question 1(c) Merit

A sec–tan integral, an area under a cubic-in-a-bracket, a kinematics chain that starts 360 m from the reference point, a trig differential equation, and an area between two trig curves.

An object’s velocity can be modelled by v(t) = 26.4∛t, where v is in m s⁻¹ and t is the time in seconds since the start of timing. Initially, the object was 360 metres from a point P.

Calculate the distance that this object has travelled from the point P when it has reached a velocity of 264 m s⁻¹.

v(t)	=	26.4∛t m s⁻¹
initial displacement 360 m from P
v	=	264 m s⁻¹

The 2024 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Write the cube root as a power, then integrate velocity to get displacement.
v = 26.4t^1/3, so s = 26.4t^4/3/(4/3) + c = 19.8t^4/3 + c.
2 Measure s from P. Where is the object when t = 0?
360 m from P, so c = 360 and s = 19.8t^4/3 + 360.
3 When does the velocity reach 264 m s⁻¹?
26.4t^1/3 = 264, so t^1/3 = 10 and t = 1000 s.
4 Put that time into s.
1000^4/3 = 10⁴, so s = 19.8 × 10 000 + 360 = 198 360 m.

Answer 198 360 m from P. The velocity is positive the whole time, so the object has been moving away from P: 198 000 m of motion added to the 360 m it started at.

What the marker wanted

Achieved Reaching s = 19.8t^4/3 + 360, with its constant.
Merit The distance, 198 360 m, from correct integration. Any sensible rounding is accepted, and units are not needed.

In our words, from NZQA’s 2024 marking schedule.

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