## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The kayak should be directed at an angle of $41.8^{\circ}$ west of north. (b) It will take 45 seconds to cross the harbor.
(a) The current will push the kayak to the east at a speed of 2.0 m/s. To go directly north, a component of the kayak's velocity should be directed to the west at 2.0 m/s. Let $\theta$ be the angle of the kayak's velocity vector which is west of north. $v_{west} = v~sin(\theta) = 2.0~m/s$ $sin(\theta) = \frac{2.0~m/s}{v}$ $\theta = arcsin(\frac{2.0~m/s}{3.0~m/s})$ $\theta = 41.8^{\circ}$ The kayak should be directed at an angle of $41.8^{\circ}$ west of north. (b) We can find the north component of the kayak's velocity vector. $v_{north} = (3.0~m/s)~cos(41.8^{\circ})$ $v_{north} = 2.24~m/s$ We can find the time to cross the harbor. $t = \frac{d}{v_{north}}$ $t = \frac{100~m}{2.24~m/s}$ $t = 45~s$ It will take 45 seconds to cross the harbor.