# 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?...

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?

*print*Print*list*Cite

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