Can anyone please explain extended question 5 e from Insight 2009 exam 2?..

[TonyZ]

in case you don't have the exam, the question is
e. Relative to the point T, find the position vector p(t) of the aircraft on the runway when the aircraft stops.

The plane touched down at point T(0,0,0)
Its velocity vector just before it touched down is -0.8 i + 4 j - 0.02e k.

[Mao]

not enough information. is the motion after touch-down modelled by anything?

[MrK]

This was left out of the question (The schools were emailed with this info):

In the question booklet, the following piece of information should have been included immediately before question 5e on p.28:

The velocity of the aircraft immediately after touchdown is   km/min, where   is the time, in minutes, after touchdown.

Mr K

[RexPP]

not enough information. is the motion after touch-down modelled by anything?

Yes. In my copy of the exam, it states that v(t) after touchdown = (-0.8i + 4j)(1-t)
So you antidifferentiate that using (0,0,0) to find p(t) and then sub in t=1, as this is when the aircraft stops.

[MrK]

Sorry, I didn't realise the equation didn't copy in.