Question #100621
Two helicopters A,B fly in a space with velocities Va=i+2j+3k and Vb=4i+2j-4k. Relative to a stationary observer at origin 0,0,0 if the helicopters fly at right angle with each other, there distance apart after 5second is
1
Expert's answer
2019-12-19T11:36:56-0500

The velocity of helicopter B relative helicopter A is


v=(4i+2j4k)(i+2j+3k)=(3i7k)ms\bold{v=(4i+2j-4k)-(i+2j+3k)=(3i-7k)}\frac{m}{s}

v=32+(7)2=7.6msv=\sqrt{3^2+(-7)^2}=7.6\frac{m}{s}

d=vt=(7.6)(5)=38 md=vt=(7.6)(5)=38\ m


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