The swimmer

The swimmer swims at a constant speed of 0.85 m/s relative to water flow. The current speed in the river is 0.40 m/s, the river width is 90 m.

a) What is the resulting speed of the swimmer with respect to the tree on the riverbank when the swimmer motion is perpendicular to the current?
b) How long does the swimmer cross the river?

Result

v =  0.939 m/s
t =  105.882 s

Solution:

v1=0.85 m/s v2=0.40 m/s s=90 m  v=v12+v22=0.852+0.42≐0.9394≐0.939 m/sv_{1}=0.85 \ \text{m/s} \ \\ v_{2}=0.40 \ \text{m/s} \ \\ s=90 \ \text{m} \ \\ \ \\ v=\sqrt{ v_{1}^2 + v_{2}^2 }=\sqrt{ 0.85^2 + 0.4^2 } \doteq 0.9394 \doteq 0.939 \ \text{m/s}
t=s/v1=90/0.85≐180017≐105.8824≐105.882 st=s / v_{1}=90 / 0.85 \doteq \dfrac{ 1800 }{ 17 } \doteq 105.8824 \doteq 105.882 \ \text{s}

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