Relative Velocity (boat traveling across a river)

[commit3d]

Im taking physics now and I am trying to study for an exam but some of the sample questions in my textbook do not give a an explanation of how to solve them...it's ridiculous. Do they think it will just come to me?!

Question:Suppose the river is moving east at 5.00km/h and the boat is traveling at 45 degrees south of east with respect to earth. Find (a) the speed of the boat with respect to Earth and (b) the speed of the boat with respect to the river if the boat's heading in the water is 60.0 degrees south of east.

there was a sample problem earlier with the same boat and it said that the speed of the boat was 10.0km/h. I am pretty sure they assumed we knew that already and is is supposed to be used. But other than that I am pretty confused. Please HELP!

 

[Anti-Meson]

Think about components of velocity... hope that starts you off.

 

[commit3d]

ok i just did the components and for the x component it is: sin(45)= x/(10km/h) so, X= 7.07km/hr
I did the same for the y component with cos(45) and got the same value. Which, i think makes sense since 45 degrees is right in the middle.

and since it says that the river flows east at 5km/h and the x component is 7.07km/h does that mean that the boat is traveling at 12.07km/hr in the x direction?

 

[Anti-Meson]

The boat is traveling 7.07 km/h + 5km/h east with respect to (wrt) the Earth. Can you see why?

Second part should follow on from this, remember 10km/h is wrt the river.

 

[commit3d]

ok yeah i did that and that's the same answer that i got. Then i did the pathagorean theorem to get the boats speed...but the answer i got is 13.99 and the book tells me 16.7km/r

 

[oluedo]

You could just use the law of sines in the resulting 15-45-120 triangle. 5/sin15 = x/sin120 and x = 16.7. Do the same for the other.

Or you could set up dual equations: YCos60 = XCos45 - 5 and YSin60 = XSin45. That should give X = 16.7 and Y=13.7. Reply if you are still stuck.