Relative Velocity while swimming

[mcintyre_ie]

Hey
I need a little help with this (B) question:

A girl wishes to siwim across a river 60m wide. The river flows with a velocity q m.s^-1 parallel to the straight banks and the girl swims at a velocity of p m.s^-1 relative to the water. In crossing the river as quickly as possible she takes 100s and is carried downstream 45m.

Find:
(I) P and Q
(II)How long will it take her to swim in a straight line back to the original starting point.

Ive done some of the basics here, which i know to be right, including the answers to part 1 - when i tried to bring it further i got answers which were obviously wrong for part 2. This is what I've got so far, any help would be appreciated:

Vr (welocity of river) = q m.s^-1
Vg/r )velocity of girl relative to river) = Vg - Vw
Vg = (pi + qj) m.s^-1

when t = 100s, d = 45i + 60j

P = 60/100 = .6m.s^-1

Q = 45/100 = .45m.s^-1

Vg/r = .6m.s^-1
Vr = .45 m.s^-1

 

[HallsofIvy]

Your calculations are correct. She will cross the river fastest by swimming straight across and allowing the river to carry her down stream. Assuming her velocity relative to the water is <p,0> and the velocity of the river is <0,q> her actual velocity is <p,q> and so displacement in 100 seconds is <100p, 100q>= <60, -45>. p, the girl's speed relative to the water is .6 m/s and the speed of the river is .45 m/s.

In order to swim "in a straight line back to the original starting point", she will have to angle sharply upstream. Take the angle her "bearing" makes with a line directly across the river to be &theta;. Then her velocity relative to the water will be <.6 cos&theta;, .6 sin&theta> so her true velocity will be <.6 cos&theta;, -.45+ .6 sin&theta>. (I'm taking positive x, here, to be back across the river.)

Here displacement vector, in t seconds, will be
<.6 cos&theta; t, -.45t+ .6 sin&theta; t>= <60, 45> That gives 2 equations for t and theta. In particular, we can write
-.45t+ .6 sin&theta; t= 45 as -.45+ .6 sin&theta;= 45/t so that
.6 sin&theta;= 45/t+ .45.

Then (.6 sin&theta;)²= (45/t+ .45)² and
(.6 cos&theta;)²= (45/t)² so, adding
(45/t+ .45)²+ (45/t)²= 0.36.

Solve that quadratic equation for t.

 

[M98Ranger]

To find the time it takes for the girl to swim back to the original starting point, we can use the formula d = rt, where d is the distance, r is the rate or velocity, and t is the time. In this case, we want to find t, so we can rearrange the formula to t = d/r.

Since the girl is swimming back to the original starting point, the distance she needs to cover is 45m (the distance she was carried downstream). The velocity she is swimming at relative to the river is still .6m.s^-1, so we can plug in these values to find t:

t = 45/.6 = 75s

Therefore, it will take the girl 75 seconds to swim in a straight line back to the original starting point.