Question

A kayaker needs to paddle north across a 90 m wide harbor. The tide is going out, creating a tidal current that flows to the east at 1.8 m/s. The kayaker can paddle with a speed of 2.8 m/s.

(a) In which direction should he paddle in order to travel straight across the harbor?
° west of north

(b) How long will it take him to cross?
s

Answer 1

here,

(a)

speed of tidal current , vt = 1.8 m/s

speed of paddling , vk = 2.8 m/s

let the angle be theta

vk * sin(theta) = vt

sin(theta) = 1.8/2.8

theta = 40 degree

the direction is 40 degree from west of north

(b)

distance , d = 90 m

time taken to cross , t = d/( vk * cos( theta))

t = 90 /( 2.8 * cos(40))

t = 41.96 s

the time taken to cross is 41.96 s