Question

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.

Answer 1

3503.72 seconds

819.96 seconds

Step-by-step explanation:

`V_r`=Velocity of river = 0.566 m/s

`V_s`=Velocity of student = 1.17 m/s

Distance to travel = 1.57 km = 1570 m

So,

Time = Distance / Speed

`(1570)/(V_s-V_r)+(1570)/(V_s+V_r)=t\\\Rightarrow t=(1570)/(1.17-0.566)+(1570)/(1.17+0.566)\\\Rightarrow t=3503.72\ s`

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

In still water

`(1570)/(V_s)+(1570)/(V_s)=t\\\Rightarrow t=(1570)/(1.17)+(1570)/(1.17)\\\Rightarrow t=2683.76\ s`

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