The velocity of the first airplane is 740 m/h at a heading of 63.4 ◦ . The velocity of the second is 650 m/h at a heading of 184◦ . How far apart are they after 1.4 h? Answer in units of m.

Answer 1

first, reduce each velocity to it's x & y components thus:

v1x=740cos(63.4)=331.2 & v1y=740sin(63.4)=661.7

v2x=650*cos(184)=-648.4 and v2y=650*sin(184)=-45.3

Second find the differences between the x velocities to find 331.2-(-648.4)= 979.6 mph

Third, find the difference between the y velocities to find 661.7- (-45.3) = 705.7 mph

Forth, convert velocity to distance. Since the time is to be 1 hour that distances are just 1*velocities.

Fifth calculate magnitude of vector difference = sqrt(979.6^2+705.7^2)=1207 miles