Question

the initial velocity of particle is 5 x 10^6 m/s in the positive x-direction. After 2.0s, the increase in its velocity is 2.5 x 10^6 m/s in the positive y-direction. Calculate the magnitude and direction of the velocity after 2.0 s.

Answer 1

Magnitude=\[\sqrt{(5\times10^6)^2+(2.5\times10^6)^2}\]
Direction=\[\tan^{-1}(2.5\times10^6/5\times10^6)\]above positive x-direction

Answer 2

apply parallelogram theorem of vectors, since the angle between the two vectors is 90 degrees(angle b/w +ve y and x axis)

magnitude = \[10^{6}\sqrt{5^{2}+2.5^{2}} = 5.59\times10^{6} m/s\]

direction = \[\tan^{-1} (2.5/5) = \tan^{-1} 0.5 = 26.565 degrees \above x- axis\]