In: Physics
A river has a steady speed of 0.650 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point.
(a) If the student can swim at a speed of 1.15 m/s in still
water, how long does the trip take? s
(b) How much time is required in still water for the same length
swim? s
(c) Intuitively, why does the swim take longer when there is a
current?
What happens if the water
is
still? The student swims a distance of 1.00 km “upstream”
at a speed of 1.15 m/s
using the simple distance/time formula
d = vt
the time for the trip is
t = d/v = 1.0×10^3m/1.15 m/s = 1000/1.15= 870
and the same is true for the trip back “downstream”. So the tot
at time for the trip is
870 s + 870 s = 1740 s = 1.74*10^3 s = 29 min
the velocity of the water with respect to the groundis
vWG = -0.650 m/s
velocity with respect to the water is
vSW= +1.15m/s.
the water and the water’s velocity with respect to the ground:
vSG=vSW+vWG = 1.15m/s-0.650m/s =0.5m/s
displacement of ∆x=1.00km
delta t=delta x/delta SG =1.00*10^3 m/0.5m/s = 2000 s
Then the student swims downstream and his velocity with respect to the water is
vSW=-1.15m/s.
velocity with respect to the ground of
vSG = vSW+vWG = -1.15m/s-0.650 m/s = -1.8m/s
time to cover a displacement of ∆x=−1.00 km is
∆t=∆x/vSG=(-1.00*10^3m)/(-1.8m/s) =555.55 s
the total time to swim upstream and then downstream is
tTotal=tup+tdown = 2000s+555.55s= 42.59 min
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(b) If the water were still the distance is
1000m/1.15m/s =869.56 s
total time is
869.56*2 = 1739.12 s
the time required in still water for the same length swim is
2555.55-1739.12 = 816.43 s = 13.6 min
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(c) when there is a current the student swims equal distances moving slowlyand moving rapidly realtive to the shore. but his average speed is the time he spends with each different speed, consider here he spends more than twice as much time with the lower spped, so his average speed is lower than it would be without a current. this depressed average speed makes the round-trip time longer when the current is flowing.