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Spoony

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## Homework Statement

Swimmer A swims due east at a constant speed of 3m/s along a straight stretch of river with neglible current, keeping a constant distance of 10m from the southern bank. A second swimmer, B, starts swimming from this bank when A is a disnace L down the river from her (i.e. L is the distance measured along the bank.) Swimmer B swims with a constant speed of 2m/s and at an angle of N60E in order to intercept A.

i) what is the velocity of b as observed from a?

ii) what is the distance L, and how long does B swim for?

iii) If a strong current is flowing how, if at all, would the above results be alterd?

## The Attempt at a Solution

I have calculated A and B's vector velocities.

Va = 3

*i*m/s

Vb = sqrt(3)

*i*+ 1

*j*

Then surely the velocity of b as observed from is just the difference between the two vectors.

Vdiff. = (3-sqrt(3))

*i*+ 1

*j*

Then i run into trouble for the next bit, surely if swimmer b's vertical component of velocity is 1m/s then no matter what distance A swims before B starts. B will always have to swim through 10m of water vertically. So if B's vertical component of velocity is 1, then he has to swim 10s before he contacts A.

Its the distance L which confuses me. B's horizontal component is sqrt(3) which is less than 3, so if B starts after A has passed B will never reach A. The distance will get progressively larger. So surely B has to start before A passes him on the river?

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