M1 trouble Help

Hello stuck on the following question:-
A coastguard station O monitors the movements of ships in a channel. At noon, the station’s radar records two ships moving with constant speed. Ship A is at the point with position vector (–5i + 10j) km relative to O and has velocity (2i + 2j) km h–1. Ship B is at the point with position vector (3i + 4j) km and has velocity (–2i + 5j) km h–1.
Given that the two ships maintain these velocities, show that they collide.

Thanks

Reply 1

Libertine

8

bowers

Hello stuck on the following question:-
A coastguard station O monitors the movements of ships in a channel. At noon, the station’s radar records two ships moving with constant speed. Ship A is at the point with position vector (–5i + 10j) km relative to O and has velocity (2i + 2j) km h–1. Ship B is at the point with position vector (3i + 4j) km and has velocity (–2i + 5j) km h–1.
Given that the two ships maintain these velocities, show that they collide.

So the position vector at time t of the ships are:
Ship A = -5i + 10j + t(2i+2j)
Ship B = 3i + 4j + t(-2i+5j)

For them to meet, i and j positions must be the same, so:
-5i + 2ti = 3i - 2ti
t = 2

If they collide, the j position vector must be the same when t=2, i.e.
10j + 2tj = 4j + 5tj
Sub in t = 2
10j + 4j = 4j + 10j
14j=14j

There is a time when both position vectors are the same, so they collide.

Reply 2

mikeyT

2

If they collide then their positions will be the same:
(-5i+10j) + (2i+2j)t= (3i+4j) +(-2i+5j)t, where t is the time in hours.
Equating the i and j coefficients:
i: -5+2t=3-2t, this gives t=2
j: 10+2t=4+5t, this gives t=2 also.

So the time t is the same for both directions i and j hence the boats collide

Reply 3

bowers

OP

Thank you very much to both of you --> now I can sleep a little bit easier :smile:

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