A swimmer maintains a speed of 0.15 m/s relative to the water when swimming directly toward the opposite shore of a straight river with a current that flows at 0.75 m/s.

Question

(a) How far downstream is the swimmer carried in 1.5 minutes?

(b) What is the velocity of the swimmer relative to an observer on shore?

This involves relative velocity, and that is about the only thing I know.

Can you help with this please? Thank you very much.

Answer 1

While the swimmer is trying to swim across the river he is also carried down
stream because of the velocity of the river. The swimmer has now two
velocities, 0.15 m/s which is the velocity of his swim and 0.75 m/s which is
the velocity of the river. The velocity vx = 0.75 m/s take him down stream
and the velocity vy = 0.15 m/s take him across the river. Since the velocity
of the river is a constant, the distance x the swimmer travels down stream is
given by
x = vxt = 0.75 x 1.5 x 60 = 67.5 m
(b)
The velocity of the swimmer relative to someone on the shore is the resultant
of 0.75 i and 0.15 j. The magnitude of this resultant is:
2 2 1 0.75 0.15 0.76 . m s
−
+ = .
The direction of the resultant is given by 2.0
0.75
15.0
tan = = =
x
y
θ .
Therefore, θ = tan-1 0.2 = 11.3o
.
The velocity of the swimmer relative to an observer on the shore = 0.75 m.s-
1
at an angle θ = 11.3o
measured counter clockwise with the shore.