Question

The position of an object with a mass of `3 kg` is given by `p(t)= t^3-3t^2+ 5`. What is the impulse applied to the object at `t= 3`?

Answer 1

`9 Kgms^-1`

Explanation:

Given, `p=t^3 -3t^2 +5`

so, velocity = `(dp)/(dt)=3t^2 -6t`

Now,impulse =change in momentum

So,`m (|(dp)/(dt)|_3 - |(dp)/(dt)|_0)=3{(3*3^2 -6*3)-(0)}=3*3=9 Kgms^-1`