Akayaker needs to paddle north across a 100-m-wide harbor. the tide is going out, creating a tidal current that flows to the east at 2.0 m/s. the kayaker can paddle with a speed of 3.0 m/s.
a. in which direction should he paddle in order to travel straight across the harbor?
b. how long will it take him to cross?

Refer to the diagram shown below.

Part a.
In order for the kayaker to travel straight north across the river, the kayak should travel at angle θ west of north.
By definition,
sinθ = 2/3   =>  θ = sin⁻¹(2/3) = 41.8°.

The kayaker should paddle at an angle of about 42° west of north.

Part b.
The resultant traveling velocity is
  v = 3 cosθ
     = 3cos(41.8°)
     = 2.236 m/s

The time required to cross the river is
t = (100 m)/(2.236 m/s)
  = 44.7 s

It will take the kayaker about 45 s to cross.

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