Answer to Question #97098 in Mechanics | Relativity for babi broke my heart.

Question

On a beautiful summer night a pensive and motionless raccon is watching a log floating down an indolent river with a velocity of 1.5 m/s, East 25 degrees North relative to the racoon. The racoon also notices that a pesky muskrats is approaching the log at a velocity of 3.75 m/s, South 40 degrees East relative to the racoon in an attempt to scare an innocent frog sleeping on the log. If the frog wakes up in time and tries to evade the muskrats by “running” along the log with a velocity of 0.3 m/s straight South relative to the log, find the frog’s velocity with respect to the racoon.

Answer 1

$$\bold{v_1}=(1.5\cos{25},1.5\sin{25})$$

$$\bold{v_2}=(-0.3,0)$$

$$\bold{V}=(1.5\cos{25}-0.3,1.5\sin{25})=(1.059,0.634)\frac{m}{s}$$

$$V=\sqrt{1.059^2+0.634^2}=1.2\frac{m}{s}$$

$$\theta=\arctan{\frac{0.634}{1.059}}=E\ 30\degree \ N$$