# Relative Motion

Hi! I have an exercise here that I can't solve. Help please.

"One of two cylists cycles in the north-west direction at a speed of 8 m/s making an angle of 53 with the west. The other cyclist cycles in north-east direction at a speed of 6m/s making an angle of 37 with the east. How does the second cyclist see the motion of first one?" (magnitude of velocity vector and direction)

Thanks........ :

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jcsd
Gold Member
As usual show us what you've done.

The obvious starting point is: What is the angle between the two cyclists?

It must be 90

jcsd
Gold Member
Yes, so what will the coponents of the relative velocity of the first cyclist (to the second cyclist)

a) along his (the second cyclist's) diretcion of travel

b) 90 degrees to his direction of travel

I've found all Vx and Vy 's and Vrelative is 10m/s to North -West. But the answer in book is 10m/s to North-east

jcsd
Gold Member
It sounds to me is what you've done is forgotten that the component of the first cyclist's velocity (relative to the second cyclist) along the the second cyclist's axis will be opposite and equal to the second cyclist's actual velocity.

Doc Al
Mentor
Azeri said:
I've found all Vx and Vy 's and Vrelative is 10m/s to North -West. But the answer in book is 10m/s to North-east
Your answer makes sense to me. Since they are separating along the east-west axis, the relative velocity (of the first with respect to the second) must have a westward component.

I use this formula: Vrelative=Vc1-Vc2

Doc Al
Mentor
Azeri said:
Vrelative=Vc1-Vc2
That is correct.