A river flows due south with a speed of `2.0 ms^-1`. A man strees a motorboat across the river , his velocity relative to the water is `4 m 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 points will he reach the opposite bank ?

Explanation

Velocity of river(i.e., speed of river w.r.t earth),
`|vec v_r| = 2 m s^-1`
velocity of boat w.r.t. river, `|vec v_(b//r)| = 4 ms^-1`
Width of river `= 800 m`
Velocity of boat w.r.t. earth `vec v_b = ?`
According to the given statement, the diagram will be as given in (Fig ).
(a) When two vectorc are acting at angle of `90^@`, their resultant can be obtained by pythagorous theorem,
`v_b = sqrt(v_(b,r)^2 + v_r^2) = sqrt(16 + 4) = sqrt(20) =45 m s^-1`
To find direction, we have
`tan theta = (v_r)/(v_(b,r))= (2)/(4) = (1)/(2) rArr theta = tan^-1 ((1)/(2))`
(b) Time taken to cross the river is
(Displacement)/velocity of boat w.r.t. river)
velocity of boat w.r.t. river is used since it is the velocity with which the river is crossed.
So the boat will sross in `(800)/(4) = 200 s`
( 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 `(2 ms^-1)`.
So the distance `AB` = Time taken xx speed of current
=`200 xx 2 = 400 m`
Hence, the boat is drifted by `400 m` away from position `A`.
.