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Q: A boater wants to cross a river that has a current of 5mph and measures .75 miles in width. If the boat can travel at 20mph through the water, what heading must it take in order to travel from A and arrive directly across the river at B?

A: I was thinking about making about making two right angle triangles back to back (each containing the same measurements) The bottom of the triangle being (.75/2=.375).

I used the 5mph as deceleration? However, I am not sure if one can do that.

So this is how I tried to solve it…but I think it is wrong!

v v0+at

v20mph + -5mph (t)

t4s

y=y0 +v0(t)+.5a(t^2)

y=0+8.94m/s(4s) +.5(2.2m/s^2)(16)

y=18.6…but that seems too big so I need help!!!

Thanks,

Tucky