## College Physics (4th Edition)

(a) The speed of the current is $1.00~m/s$ (b) The boy's speed relative to a friend on the riverbank is $1.12~m/s$
(a) We can find the time it takes for the boy to swim across the river: $t = \frac{25.0~m}{0.500~m/s} = 50.0~s$ In $50.0~s$, the current carried the boy $50.0~m$ downstream. We can find the speed of the current: $v = \frac{50.0~m}{50.0~s} = 1.00~m/s$ The speed of the current is $1.00~m/s$ (b) We can find the boy's speed relative to the riverbank: $\sqrt{(0.500~m/s)^2+(1.00~m/s)^2} = 1.12~m/s$ The boy's speed relative to a friend on the riverbank is $1.12~m/s$