Question #95331
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?
1
Expert's answer
2019-09-27T09:47:06-0400

The velocity of boat 1 relative to boat 2 is:


v1v2=(2.15sin47.0°,2.15cos47.0°)\bold{v_1 - v_2}= (2.15\sin{47.0°}, 2.15\cos{47.0°})v1=(0,0.775)\bold{v_1}= (0, 0.775)

(0,0.775)v2=(2.15sin47.0°,2.15cos47.0°)(0, 0.775)-\bold{ v_2}= (2.15\sin{47.0°}, 2.15\cos{47.0°})

v2=(1.572,0.691)\bold{ v_2}=(-1.572,-0.691)

v2=1.5722+0.6912=1.72msv_2=\sqrt{1.572^2+0.691^2}=1.72\frac{m}{s}

The direction:


θ=arctan0.6911.572=23.7°south of west\theta =\arctan{\frac{0.691}{1.572}}=23.7\degree south\ of\ west





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Comments

Patience
06.10.19, 16:04

I really need help in physics ad math

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