Question

A river flows due south with a speed of $2.0m{s}^{-1}$. A man steers a motorboat across the river, his velocity relative to the water is $4m{s}^{-1}$ due east. The river is $800$ m wide.
a. What is his velocity (magnitude and direction) relative to the earth?
b. How much time is required to cross the river?
c. How far south of his starting point will he reach the opposite bank?

Answer 1

Velocity of river (i.e., speed of river w.r.t. earth),

$\overrightarrow{v_{r/e}}=2m{s}^{-1}$

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

Width of river = $800m$

Velocity of boat w.r.t. earth $\overrightarrow{v_{b/e}}=?$

According to the given statement, the diagram will be as given in Fig.

a. When two vectors are acting at angle of 900, their resultant can be obtained by pythagorous theorem,

${\overrightarrow{v}}_{b/e}=\sqrt{v_{b/r}^2+v_{r/e}^2}=\sqrt{16+4}=45m{s}^{-1}$

To find direction, we have

$tan\theta =\frac{v_{r/e}}{v_{b/r}}=\frac{2}{4}=\frac{1}{2}\Rightarrow \theta =ta{n}^{-1}\left(\frac{1}{2}\right)$

b. Time taken to cross the river is

$\frac{Displacement}{Velocityofboatw.r.triver}$

Velocity of boat w.r.t. river is used since it is the velocity with which the river is crossed.

So the boat will cross in $\frac{800}{4}=200s$

c. The desired position on other side is $A$, but due to the current of river, boat is drifted to position $B$. To find out this drift, we need time taken in all to cross the river ($200$ s) and speed of current ($2m{s}^{-1}$).

So the distance $AB$ = Time taken $\times$ Speed of current

$=200\times 2=400m$

Hence, the boat is drifted by $400$ m away from position A.