# The velocity of an object with a mass of 2 kg is given by v(t)= 3 t^2+ 2 t +8. What is the impulse applied to the object at t= 4 ?

Jan 15, 2016

The impulse at $t = 4$ is $52 k g m {s}^{-} 1$

#### Explanation:

Impulse is equal to the rate of change of momentum:

$I = \Delta p = \Delta \left(m v\right)$.

In this instance the mass is constant so $I = m \Delta v$.

The instantanteous rate of change of the velocity is simply the slope (gradient) of the velocity-time graph, and can be calculated by differentiating the expression for the velocity:

$v \left(t\right) = 3 {t}^{2} + 2 t + 8$
$\frac{\mathrm{dv}}{\mathrm{dt}} = 6 t + 2$

Evaluated at t = 4, this gives $\Delta v = 26 m {s}^{-} 1$

To find the impulse, then, $I = m \Delta v = 2 \cdot 26 = 52 k g m {s}^{-} 1$