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Describing motion in 2D using vectors

Two ships P and Q are moving with constant velocities. The velocity of P is (9i-2j)km h¯¹ and the velocity of Q is (4i+8j) km h¯¹.
a) Find the direction of motion of P, giving your answer as a bearing to the nearest degree.

When t=0 the position vector of P is (9i+10j)km and the position vector of Q is (i+4j) km. At time t hours, the position vectors of P and Q of P and Q are p km and q km respectively.
b) Find an expression for
I) p in terms of t
II) q in terms of t
C) Hence show that at time t hours QP=(8+5t)i+(6-10t)j
D)Find the values of t when ships are 10 km apart
(edited 4 years ago)
Reply 1
Please try to attempt the question first and then upload your workings
Reply 2
Original post by vbzl
Please try to attempt the question first and then upload your workings

I have attempted to answer it but I'm stuck on the last question please may you help me
Reply 3
Original post by vbzl
Please try to attempt the question first and then upload your workings

I have tried to attempt the question I'm just stuck on the last part of the question I'm not sure how to find the value of t
Reply 4
You have the displacement vector between the two ships and you want the distance of 10km between the two ships.

The Pythagoras Theorem might help here and you should end up with a quadratic to solve.
Reply 5
Original post by simon0
You have the displacement vector between the two ships and you want the distance of 10km between the two ships.

The Pythagoras Theorem might help here and you should end up with a quadratic to

I've tried to do that but I keep ending up with imaginary numbers.
Reply 6
Original post by Quasar205
I've tried to do that but I keep ending up with imaginary numbers.

Length of QP=10
QP^2=((8+5t)^2 +(6-10t)^2)^0.5
Just solve it.
Reply 7
Original post by Quasar205
I've tried to do that but I keep ending up with imaginary numbers.

Did you obtain:

125t240t+100=100? \displaystyle 125 t^{2} - 40t + 100 = 100 \, ?

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