Question

A swimmer heads directly across a river, swimming at 1.9 m/s relative to the water. She arrives at a point 50 m downstream from the point directly across the river, which is 79 m wide.
(a) What is the speed of the river current?
m/s

(b) What is the swimmer's speed relative to the shore?
m/s

(c) In what direction should the swimmer head so as to arrive at the point directly opposite her starting point?
° upstream from straight ahead

Answer 1

tan A = 50m/79m = 0.63291
A = 32.3o East of North = 57.7o CCW.

b. Vs = 1.9m/sin57.7 = 2.25 m.[57.7o] = Velocity of swimmer.

Vc = 2.25m/s[57.7o] - 1.9[90o]

X = 2.25*cos57.7 - 1.9*cos90=1.20 m/s.
Y=2.25*sin57.7 - 1.9*sin90 = 0 m/s.

b. Vc = 1.20 + i0 = 1.20 m/s =Velocity of the current.

c. 32.3oWest of North.


Vc = X/cosB = 1.20/cos0.0878=13.67m/s.

Answer 2

A man can swim directly across a stream of width 300 m in 4 minutes, when there is no current and in 5 minutes when there is current. Find velocity of the current.