Velocity of a particle changes from `(3hati +4hatj)m//s` to `(6hati +5hatj)m//s` 2s. Find magnitude and direction of average acceleration.

Explanation

Given, velocities of the particle
`v_(1) = 3hati +4hatj` and `v_(2) = 6hati +5hatj`
Change is velocity
`Deltav_(x) = (v_(2) x - v_(1)x) hati = (6-3)hati = 3hatj`
`Deltav_(y) = (v_(2)y - v_(1))hatj = (5-4)hatj = hatj`
`:. Deltav = Deltav_(x) +Deltav_(y) = 3hati +hatj`
`:.` Average acceleration
`a_(av) = (Deltav)/(Deltat) = (3hati +hatj)/(2) ( :' t = 2s)`
`=1.5 hati +0.5 hatj`
Direction of averae acceleration
`tan theta = (a_((av)y))/(a_((av)x)) =(0.5)/(1.5) =(1)/(3)`
`rArr theta =tan^(-1) ((1)/(3)) ~~ 18.43^(@)` with x-axis