# 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 ?

Mar 11, 2018

$9 K g m {s}^{-} 1$

#### Explanation:

Given, $p = {t}^{3} - 3 {t}^{2} + 5$

so, velocity = $\frac{\mathrm{dp}}{\mathrm{dt}} = 3 {t}^{2} - 6 t$

Now,impulse =change in momentum

So,$m \left(| \frac{\mathrm{dp}}{\mathrm{dt}} {|}_{3} - | \frac{\mathrm{dp}}{\mathrm{dt}} {|}_{0}\right) = 3 \left\{\left(3 \cdot {3}^{2} - 6 \cdot 3\right) - \left(0\right)\right\} = 3 \cdot 3 = 9 K g m {s}^{-} 1$