# Boating problem

tucky
Hi everyone…it feels like it has been a while but I have another question? This is dealing with vectors.

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

Loren Booda
Try vector representation. (The given width, you should recognize, is extraneous information toward finding the heading for linear flow.)

Think of a right triangle with legs of 20 mph and 5 mph, perpendicular and parallel to the river's flow. Find the resultant direction needed to offset the river's velocity.