# The velocity of an object with a mass of 3 kg is given by v(t)= - t^2 +4 t . What is the impulse applied to the object at t= 5 ?

Jan 3, 2016

The impulse of an object is associated to a change on its linear momentum, $J = \Delta p$.
Let us calculate it for $t = 0$ and $t = 5$.

#### Explanation:

Let us suppose that the object starts its motion at $t = 0$, and we want to calculate its impulse at $t = 5$, i.e. the change of linear momentum it has experienced.

Linear momentum is given by: $p = m \cdot v$.

• At $t = 0$, linear momentum is:
$p \left(0\right) = m \cdot v \left(0\right) = 3 \cdot \left(- {0}^{2} + 4 \cdot 0\right) = 0$

• At $t = 5$, linear momentum is:
$p \left(5\right) = m \cdot v \left(5\right) = 3 \cdot \left(- {5}^{2} + 4 \cdot 5\right) = - 15 \text{ kg" cdot "m/s}$

So impulse finally is given by:

$J = \Delta p = p \left(5\right) - p \left(0\right) = \left(- 15\right) - \left(0\right) = - 15 \text{ kg" cdot "m/s}$

The negative sign just means that the object is moving backwards.

P.S.: the vectorial expression is $\vec{J} = \Delta \vec{p}$, but we have assumed that object moves only on one direction, and we just take in account the modulus of the magnitudes.