Mo and Dave are in a canoe crossing a very wide stream. Their velocity is represented by a vector, v = <0 , 15>. The stream which runs east to west at 5 mph can be represented by a vector, v = <-5 , 0>. What is the actual representation of both of them in their canoe? What is their actual speed?
Thankyou!
2007-04-28 12:06:16 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I assume you mean their velocity in still water would be
v = <0, 15>. Otherwise, if v is their velocity in the stream you are already done.
Assuming the former, the actual representation of both of them in their canoe is
<0, 15> + <-5, 0> = <-5, 15>.
Their actual speed is:
s = √[(-5)² + 15²] = √(25 + 225) = √250 = 5√10 ≈ 15.811 mph
2007-04-28 12:19:36 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
Well, they are travleing 15mph north and the stream is running 5 mph west so draw a triangle where the hypotonous is the resultant vector which will be sqrt((-5)^2 + 15^2), using pythagorean theorem, and the angle they travel in will be arctan(-5/15)
2007-04-28 12:20:42 · answer #2 · answered by Johnny 3 · 0⤊ 0⤋