# An object with a mass of 4 kg is traveling at 3 m/s. If the object is accelerated by a force of f(x) = 2x^2 -x +3  over x in [0, 9], where x is in meters, what is the impulse at x = 2?

Jun 30, 2016

$3.04 \frac{k g m}{s}$

#### Explanation:

Given

$m - \text{Mass of the oject} = 4 k g$

$u \to \text{Initial velocity} = 3 \frac{m}{s}$

$\text{Force } , f \left(x\right) = 2 {x}^{2} - x + 3$

So $\text{Acceleration } a \left(x\right) = \frac{1}{m} \cdot f \left(x\right)$

Again we know $a \left(x\right) = v \frac{\mathrm{dv}}{\mathrm{dx}}$

So
$v \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{m} f \left(x\right) = \frac{1}{4} \left(2 {x}^{2} _ x + 3\right)$

$\implies \int v \mathrm{dv} = \frac{1}{4} \int \left(2 {x}^{2} - x + 3\right) \mathrm{dx}$

$\implies {v}^{2} / 2 = \frac{1}{4} \left(2 \cdot {x}^{3} / 3 - {x}^{2} / 2 + 3 x\right) + c$

$\implies {v}^{2} = {x}^{3} / 3 - {x}^{2} / 4 + \frac{3}{2} x + c$

$\text{Given at } x = 0 , v = 3$

we get c=9

So now

$v \left(x\right) = \sqrt{{x}^{3} / 3 - {x}^{2} / 4 + \frac{3}{2} x + 9}$

When x = 2 velocity becomes

$v \left(2\right) = \sqrt{{2}^{3} / 3 - {2}^{2} / 4 + \frac{3}{2} \cdot 2 + 9}$

$= \sqrt{\frac{8}{3} - \frac{1}{2} + 3 + 9} = \sqrt{14.16} \frac{m}{s} = 3.76 \frac{m}{s}$

Hence momentum of the object at x=2 is

$\text{Momentum "= "mass"xx"velocity}$
$= 4 \times 3.76 = 15.04 \frac{\text{kgm}}{s}$

So impulse which is change in momentum during the time to reach at $x = 2$ from $t = 0$ will be

$I = \left(15.04 - 3 \cdot 4\right) \text{kgm/"s=3.04kgm"/} s$