Question

A man crosses a 320 m wide river perpendicular to the current in 4 min. if in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in m/min is?

(a) 30

(b) 40

(c) 50

(d) 60

plz answer the above question with an appropriate solution.

Answer 1

Let the velocity of swimmer = V(s) 

Let the velocity of current    = V(w)

width of river = 320 m

time taken to cover the river = 4 minute

the swimmer swims in -ve direction so that the resultant velocity would be perpendicular to the river velocity  .

so the resultant velocity or V(sw) has two components 

V(sw) sinФ in the perpendicular 

V(sw) cosФ in the base direction 


As we know V(sw) sin Ф = V(w) 

Also V(sw) cos Ф = V(s) 


And 

t = 320 / V(sw)cosФ

4 = 320 / V(sw)cosФ


In the question it is given that 

the 5/3V(w) = V(sw) 

So , 

we can write 

4 = 320 / (5/3) V(w) x 4/ 5

V(w) = 320 x 3 / 4 x 4 

V(w) = 20 x 3 

V(w) = 60 m/min

So option d) is correct 

Answer 2

Answer:d:60

Explanation:

V=speed of man

U=speed of river

V=5/3 u

Account to the formula,

t = d/(√v^2-u^2)

4=320/(√(5/3 u)^2 - u^2)

On solving we get,

D=60