Answer to Question #95331 in Classical Mechanics for Jake

Question

As two boats approach the marina, the velocity of boat 1 relative to boat 2 is 2.15 m/s in a direction 47.0° east of north. If boat 1 has a velocity that is 0.775 m/s due north, what is the velocity (magnitude and direction) of boat 2?

Answer 1

The velocity of boat 1 relative to boat 2 is:

$$\bold{v_1 - v_2}= (2.15\sin{47.0°}, 2.15\cos{47.0°})$$ $$\bold{v_1}= (0, 0.775)$$

$$(0, 0.775)-\bold{ v_2}= (2.15\sin{47.0°}, 2.15\cos{47.0°})$$

$$\bold{ v_2}=(-1.572,-0.691)$$

$$v_2=\sqrt{1.572^2+0.691^2}=1.72\frac{m}{s}$$

The direction:

$$\theta =\arctan{\frac{0.691}{1.572}}=23.7\degree south\ of\ west$$