A man crosses a 320 tn wide river perpen-dicular to the current in 4 m

1.	A man crosses a 320 tn wide river perpen-dicular to the current in 4 min. If in stillwater he can swim with a speed 5/3 timesthat of the current, then the speed of thcurrent in m/min is43.t 4
Answer» 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) sinФ = V(w) Also V(sw) cosФ = 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 hit like if you find it useful

A man crosses a 320 tn wide river perpen-dicular to the current in 4 min. If in stillwater he can swim with a speed 5/3 timesthat of the current, then the speed of thcurrent in m/min is43.t 4

Answer»

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) sinФ = V(w)

Also V(sw) cosФ = 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

hit like if you find it useful

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