In an anniversary celebration of Marilyn Bell's 1954 crossing of Lake Ontario a swimmer set out from the shores of New York and maintained a velocity of 4m/s [N]. As the swimmer approached the Ontario shore, she encountered a cross current of 2m/s [E 25deg S]. Find her velocity with respect to the crowd observing from the beach.
The Attempt at a Solution
Firstly, am I to understand that this is a right-angle triangle? When representing it graphically, it certainly does not look like a right triangle:
However, when I endeavor to solve this by converting from polar to cartesian co-ordinates, it seems that I have to assume a right-triangle.
Let S be the swimmer, W be the water, and G the ground.
sVw = 4 m/s [N] = (4, 90*)
wVg = 2 m/s [E25*S] = (2, -335*)
sVg = ?
.: sVg = sVw + wVg
= [0, 4] + [1.8, 0.84]
= [1.8, 4.84]
= (5.2, 69.5*)
I am not very confident in my answer. For starters, I am not supposed to really solve this using polar-cartesian conversion, but I was at a standstill when attempting another solution. Any light shed on a solution for this would be much appreciated.
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