A particle with velocity `vec(v)_(0)=-2hat(i)+4hat(j)` (in meters per second) at `t = 0` undergoes a constant acceleration a of magnitude `a =3m//s^(2)` at an angle `theta=127^(@)` from the positive direction of the x axis. What is the particle's velocity `vec(a)` at `t=5` sec, in unit vector notation ?

Explanation

We know that `v-v_(0)+at`
now `v_(x)-v_(0 x)+a_(x)t` and `v_(y)-v_(0 y)+a_(y)t`
`a_(x)=a cos theta =(3 m//s^(2))(cos 127^(@))=1.80 m//s^(2)`
`a_(y)=a sin theta = (3 m//s^(2))(sin 127^(@))=+ 2.40 m//s^(2)`
at time `t=5` sec
`v_(x)=-2 m//s+(-1.80 m//s^(2))(5 sec)=-11 m//s`
`v_(y)=4 m//s+ (2.40 m//s^(2))(5 sec)=16 m//s`
Thus, at `t=5` sec,
`vec(v)=(-11 m//s)hat(i)+(16 m//s) hat(j)`