The force on a particle of mass `10g` is `(hati 10+hatj 5)`N If it starts from rest what would be its position at time `t=5s` ?

Answer

` vec(r) =(hati 12500 + hatj 6250) m` .

Explanation

From `vec (F) = (hati 10 + hatj5) N,` we have
`F_(x) = 10N, F_(y) = 5 N`
`:. a xx=(F_(x))/(m)= (10)/(0.01) = 1000 m//s^(2)`
As acceleration along x-axis is constant
`:. x = u_(x) t + (1)/(2) a_(x) t^(2) = 0 + (1)/(2) xx1000 xx5^(2)`
`= 12500m`
Similarly `a_(y) = (F_(y))/(m) = (5)/(0.01) = 500m//s^(2)`
and `y = u_(y) t + (1)/(2) a_(y) t^(2) = 6250m`
Hence position of particle at `t = 5 s` is
`vec(r) = (hati 12500 + hatj 6250)m` .