Question

A kayaker needs to paddle north across a 100 -m wideharbor. The tide is going out, creating a tidal current that flowsto the east at 2.0 m/s. The Kayaker can paddle with a speed of3.0m/s.
a. In which direction shoule he paddle in order to travel straightacross the harbor ?
b. How long will it take him to cross ?

Answer 1

Answer:

a) 48.2 degrees north of west

b) 44.72 s

Explanation:

In order to travel straight across the harbor, his horizontal component velocity must be the same magnitude and opposite direction as the tidal velocity. So it should be 2m/s west ward

a. The direction would be:

cos α = 2/3

$\alpha = cos^{-1} 2/3 = 0.841 rad = 0.841*180/\pi = 48.2^0$ north of west

b. His vertical component of velocity would be

$v_v = \sqrt{v^2 - v_h^2} = \sqrt{3^2 - 2^2} = \sqrt{5} = 2.24m/s$

So the time it takes for him to cross the 100m river at the rate of 2.24 m/s is

100 / 2.24 = 44.72 s