Answer:
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
[tex]x(t)=3tsin(\theta )[/tex]
[tex]y(t)=3tcos(\theta )[/tex]
where [tex]\theta[/tex] 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) [tex] 3tsin(\theta )=-100+2t[/tex]
2) [tex]3tcos(\theta )=150[/tex]
Operating on 2)
[tex]cos(\theta )=\frac{50}{t}[/tex]
Since
[tex]sin(\theta )=\sqrt{1-cos^2(\theta )}[/tex]
replacing in 1) we have
[tex]-3t\sqrt{1-\left ( \frac{50}{t} \right )^2}=-100+2t[/tex]
squaring both sides
[tex]9t^2(1-(\frac{50}{t})^2)=(-100+2t)^2\Rightarrow 9t^2-22500=10000-400t+4t^2[/tex]
and we obtain the quadratic equation
[tex]t^2+80t-6500=0[/tex]
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
[tex]cos(\theta )=1[/tex]
and [tex]\theta=0[/tex]
a)
The angle relative to north you should swim is 0 degrees, that is, you should swim perpendicular to the shoreline
b)
If we do not take into account the water friction, the distance towards east from your starting point is 0 mt
