# Boating problem

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

Related Introductory Physics Homework Help News on Phys.org
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.

HallsofIvy