A river `600 m` wide wide flows at the rate of `8 km h^(-1). Still water, wishes to cross the river straight (i) Along what direction must be strike ? What will be his resultant velcity ? (ii) How much time he will take to cross the river ?

Explanation

Refer to fig, 2 9c ) . 49,
.
velocity of river, | vec v-r| = |vec (OP)| = 8 km h^(-1)`
velocity of swimmer in still water,
` | vec v_s | = | vec (OB) | = 10 km h^-1)`
9i0 The swimmer will cross the river straight if the resultant velocity ` vec v` of velocity of river and velcoty of swimmer ` vec _s` acts along ` vec (OP)`, i.e., perpendicular to the bank of ricer. It will be so it the swimmer moces alon g` vec (OB)` where ` /_BOC= theta`. In right angled triangle ` OBC`
` sin theta = (BC)/(OB) = v_r/ v_s = 8/ (18) = 0.8= sin 53 ^@ 8'`
or ` theta = 53^@ 8'`
Therefore, the angle at which the swimmer is to swim ` = 90^@ + 53^@ 8' = 143^@ 8'` with the river stream.
(ii) Resultant velocity of at swimmer ltbr. ` v= sqrt(v_s^2-v_e^2) =sqrt 910^2 -8^2 )= 6 km h^(-1)`
`= 5/3 ms^(-1)`
(iii) Time taken to cross the river,
` t =(width of ricer 0/v = (600 m) /(5//3 ms^(-1)) = 36 0 s = 6 min`.