A motorboat heads due east at 10.7 m/s across a river that flows toward the south at a speed of 4.9 m/s.
a) What is the magnitude of the resultant velocity relative to an observer on the shore? Answer in units of m/s.
b) What is the angle from the original heading (with counterclockwise positive) of the boat’s displacement? Answer in units of ◦ .
c) If the river is 1373 m wide, how long does it take the boat to cross?
The resultant velocity is given as the resultant vector magnitude as:
`V_R = sqrt(4.9^2 + 10.7^2) = 11.8 m/s`
The direction of this vector is given as
`theta = tan^(-1) (4.9/10.7)` = 24.6 degrees (in South-East direction or 24.6 degrees towards south from east direction)
Since the counter-clockwise direction is positive, resultant is -24.6 degrees from the original heading.
The time taken to cross the river can be calculated as the ratio of distance to the velocity (component of resultant velocity in that direction)
thus, `t = (distance)/V_e = 1373/(V_R cos theta)`
= 128.4 sec
Hope this helps