The force on a particle of mass 10 g is `(veci10 + vecj5)N`. If it starts from rest what would be its position at time `t=5s?

Explanation

We have `F_x=10 N` giving,
`a_x=F_x/m=(10N0/(0.01 kg)= 1000 m/s^2`.
As this is a case of constnt acceleration in x-direction.
`x=u_x+1/2 a_x t^2 = 1/2 xx1000 m/s^2xx)5s)^2`
`=12500 m`
Similarly, `a_y=F_y/m=(5N)/(0.01 kg)=500 m/s^2`
and y=6250 m
Thus, the positon of the particle at t=5 s is
`vecr=(veci 12500+vecj 6250)m`.