Question

8. Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 33 m/s, what the magnitude of bigger vector?

Answer 1

let magnitude of first vector is v.

second vector magnitude will be 2*v.

as they are separated by 90 degrees,

magnitude of their resultant will be sqrt(magnitude of first vector^2+magnitude of second vector^2+2*magnitude of first vector*second vector magnitude*cos(angle ))

=sqrt(v^2+(2*v)^2+2*v*(2*v)*cos(90))

=sqrt(5*v^2)

given magnitude of resultant is 33 m/s

==>sqrt(5*v^2)=33

==>v=sqrt(33^2/5)

=14.758 m/s

then magnitude of the bigger vector=2*v=29.516 m/s