A river has a steady speed 0.500 m/s. A student swims upstream a distance of1.00 km and swims back to the starting point. a) If the student can swim ata speed of 1.20 m/s in still water, how long does the trip take? b) How muchtime is required in still water for the same length swim? c) Intuitively, whydoes the swim take longer when there is a current?

c) Intuitively, it takes longer to complete the trip when there is current because, the swimmer spends much more time swimming at the net low speed (0.7 m/s) than the time he spends swimming at higher net speed (1.7 m/s).

Explanation:

The problem deals with relative velocities.

Call Vr the speed of the river, which is equal to 0.500 m/s
Call Vs the speed of the student in still water, which is equal to 1.20 m/s
You know that when the student swims upstream, Vr and Vs are opposed and the net speed will be Vs - Vr
And when the student swims downstream, Vr adds to Vs and the net speed will be Vs + Vr.

Now, you can state the equations for each section:

distance = speed × time
upstream: distance = (Vs - Vr) × t₁ = 1,000 m
downstream: distance = (Vs + Vr) × t₂ = 1,000 m

Part a). To state the time, you substitute the known values of Vr and Vs and clear for the time in each equation:

(Vs - Vr) × t₁ = 1,000 m
(1.20 m/s - 0.500 m/s) t₁ = 1,000 m⇒ t₁ = 1,000 m / 0.70 m/s ≈ 1429 s
(1.20 m/s + 0.500 m/s) t₂ = 1,000 m ⇒ t₂ = 1,000 m / 1.7 m/s ≈ 588 s
total time = t₁ + t₂ = 1429s + 588s = 2,017s
Convert to minutes: 2,0147 s ₓ 1 min / 60s ≈ 33.6 min

Part b) In this part you assume that the complete trip is made at the velocity Vs = 1.20 m/s

time = distance / speed = 1,000 m / 1.20 m/s ≈ 833 s ≈ 13.9 min

Part c) Intuitively, it takes longer to complete the trip when there is current because the swimmer spends more time swimming at the net speed of 0.7 m/s than the time than he spends swimming at the net speed of 1.7 m/s.