Mathematics: A300 m wide river flows 2 m/s due east (downstream). your friend is in the river 150 m north and 100 m west of you (you are on the shoreline). your friend drifts at the speed of the river, but you can swim at 3 m/s relative to the water. assuming you start swimming immediately, a) at what angle (relative to north) should you swim to reach your friend in the shortest amount of time? b) how far downstream (i.e., east of your original location) are you when you reach your friend?

C) 92 DAYSStep-by-step explanation:V=S/tt=? S=48 360 000milesV=22000mpht=V/S...

A300 m wide river flows 2 m/s due east (downstream).

s9090484

03.01.2022

a) 0°

b ) 0 mt

Step-by-step explanation:

We are going to deprecate the water friction.

The situation is depicted in the picture attached.

Since your friend travels at a speed of 2m/seg parallel to the shoreline, after t seconds, the coordinates of your friend P(x(t),y(t)) are

P = (-100+2t, 150)

Your coordinates are

$x(t)=3tsin(\theta )$

$y(t)=3tcos(\theta )$

where $\theta$ is the angle relative to north and r(t) is the distance from Q (your position at time t) to the origin.

You will reach your friend when P=Q, that is

1) $3tsin(\theta )=-100+2t$

2) $3tcos(\theta )=150$

Operating on 2)

$cos(\theta )=\frac{50}{t}$

Since

$sin(\theta )=\sqrt{1-cos^2(\theta )}$

replacing in 1) we have

$-3t\sqrt{1-\left ( \frac{50}{t} \right )^2}=-100+2t$

squaring both sides

$9t^2(1-(\frac{50}{t})^2)=(-100+2t)^2\Rightarrow 9t^2-22500=10000-400t+4t^2$

and we obtain the quadratic equation

t^2+80t-6500=0

whose solutions are

t = -130 and t =50

As t represents time, the suitable solution is t = 50

When t = 50, replacing in 2) we get

$cos(\theta )=1$

and $\theta=0$

The angle relative to north you should swim is 0 degrees, that is, you should swim perpendicular to the shoreline

If we do not take into account the water friction, the distance towards east from your starting point is 0 mt

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