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A river has a steady speed of 0.566 m/s. A student swims upstream a distance of 1.57 km and returns (still swimming) to the starting point. If the student can swim at a speed of 1.17 m/s in still water, how long does the trip take? Answer in units of s. Compare this with the time the trip would take if the water were still; i.e., what is the time in the river minus the time in still water? Answer in units of s.

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Answer:

3503.72 seconds

819.96 seconds

Explanation:

[tex]V_r[/tex]=Velocity of river = 0.566 m/s

[tex]V_s[/tex]=Velocity of student = 1.17 m/s

Distance to travel = 1.57 km = 1570 m

So,

Time = Distance / Speed

[tex]\frac{1570}{V_s-V_r}+\frac{1570}{V_s+V_r}=t\\\Rightarrow t=\frac{1570}{1.17-0.566}+\frac{1570}{1.17+0.566}\\\Rightarrow t=3503.72\ s[/tex]

Time taken by the student to complete the trip is 3503.72 seconds

In still water

[tex]\frac{1570}{V_s}+\frac{1570}{V_s}=t\\\Rightarrow t=\frac{1570}{1.17}+\frac{1570}{1.17}\\\Rightarrow t=2683.76\ s[/tex]

The difference in time between moving water and still water is 3503.72-2683.76 = 819.96 seconds

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