Solving Vectors: Boater Needs to Cross River at 20mph

[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.

 

[HallsofIvy]

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

Did you really think about this? In the first place the problem says nothing about "acceleration" or "deceleration". In the second, 5 mph is a speed, not an acceleration.

I don't see any reason to draw TWO right triangles. Draw a line directly across the river from A to B- that represents the actual course of the boat. Draw a second line perpendicular to that- straight up the river and mark its length as 5 (mph): that represents the river current.

Now draw the hypotenuse of that triangle and mark its length as 20 (mph): that represents the speed of the boat (Loren Booda had a slight error when he said "legs of 20 mph and 5 mph"). Since you are only asked for the angle, you can use arcsin(5/20).