A kayaker needs to paddle north across a 85 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 3.4 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?

(a) To travel straight across the harbor, which is to the north, the kayaker must compensate for the tidal current flowing to the east. The direction in which the kayaker should paddle can be determined using vector addition. The kayaker's velocity (v_k) relative to the water should be aimed northwest, in such a way that the resultant vector (v_r) of the current’s velocity (v_c) and the kayaker's velocity is directed north. To find the angle (θ) at which the kayaker must paddle, we can use the following relationship, where θ is the angle west of north that the kayaker must aim to counteract the current:

tan(θ) = v_c / v_k

Given v_c = 2.0 m/s and v_k = 3.4 m/s, we can solve for θ:

tan(θ) = 2.0 m/s / 3.4 m/s θ = arctan(2.0/3.4) θ ≈ 30.6 degrees

So, the kayaker should paddle at an angle of about 30.6 degrees west of north to go straight across the harbor.

(b) The time it takes for the kayaker to cross the harbor is determined by the component of his velocity that is directed across the harbor (northward), which is the same in water as it would be if there were no current, given that he is adjusting his direction to compensate for the current. This northward component is the same as his paddling speed, given that the result of his paddling direction and the current results in a straight path northward. The time (t) to cross the harbor can be found using the formula:

t = distance / speed

The distance to be crossed is 85 m, and the kayaker's paddling speed across the harbor is the magnitude of his velocity, which is 3.4 m/s. So,

t = 85 m / 3.4 m/s t ≈ 25 seconds

It will take approximately 25 seconds for the kayaker to cross the harbor.