Question

a. In which direction should the motorboat head in order to reach a point on the opposite bank directly east from the starting point? The boat's speed relative to the water remains $4m{s}^{-1}$.
b. What is the velocity of the boat relative to the earth?
c. How much time is required to cross the river?

Answer 1

As the boat has to reach exact opposite end to the point of start, the boat has to start (velocity $4m{s}^{-1}$) at an angle aiming somewhat upstream. Taking into count the push given by the current,

Velocity of boat w.r.t river, ${\overrightarrow{v}}_{b/r}=\overline{OA}=4m{s}^{-1}$

Velocity of river w.r.t Earth, ${\overrightarrow{v}}_{r/e}=\overline{AB}=2m{s}^{-1}$

Velocity of boat w.r.t. Earth, ${\overrightarrow{v}}_{b/e}m{s}^{-1}=\overline{OB}=?$

a. To find the direction of the boat in which the boat has to go, we need to find angle $\theta$.

From $△OBA,sin\theta =\frac{AB}{OA}=\frac{2}{4}=\frac{1}{2}\Rightarrow \theta ={30}^o$

Hence, the motorboat has to head at ${30}^o$ north of east.

b. To find the velocity of boat w.r.t. Earth, we can use pythagorous theorem again.

From $△OBA$, we have

$v_{b/r}^2=v_{b/e}^2+v_e^2\Rightarrow v_{b/e}^2=v_{b/r}^2-v_e^2$

$\Rightarrow v_{b/e}^2=\sqrt{v_{b/r}^2+v_e^2}=\sqrt{4^2-2^2}=2\sqrt{3}m{s}^{-1}$

c. Time taken to cross the river is

$\frac{Widthofriver}{Velocityofboatw.r.tEarth}$

$\Rightarrow t=\frac{800}{2\sqrt{3}}=\frac{400\sqrt{3}}{3}s$