# An object with a mass of 5 kg is traveling at 2 m/s. If the object is accelerated by a force of f(x) = x^2 -x +3  over x in [1, 9], where x is in meters, what is the impulse at x = 8?

Feb 19, 2016

$v \left(x\right) = \sqrt{\frac{2}{5} \left(\frac{1}{3} {x}^{3} - \frac{1}{2} {x}^{2} + 3 x + 10\right)}$
$v \left(8\right) = \sqrt{\frac{2}{5} \left(\frac{1}{3} {\left(8\right)}^{3} - \frac{1}{2} {\left(8\right)}^{2} + 38 + 10\right)} = 8.31 \frac{m}{s}$

#### Explanation:

This is a tricky question, because acceleration is given as a function of x rather the parameter t... In any case write the impulse equation:
$F \mathrm{dt} = d \left(m v\right) = m \mathrm{dv}$
So first thing we do it convert this to be in terms of $x$. From above:
$F \left(x\right) = m \frac{\mathrm{dv}}{\mathrm{dt}}$ re-write in terms of x
$F \left(x\right) = m \frac{\textcolor{red}{{d}^{2} x}}{\mathrm{dt}} ^ 2$
we have a differential equation of the form:
$\frac{{d}^{2} x}{\mathrm{dt}} ^ 2 = \frac{1}{m} F \left(x , \frac{\mathrm{dx}}{\mathrm{dt}}\right)$
Let v = dx/dt;
using chain rule $\frac{\mathrm{dv}}{\mathrm{dt}} = \frac{\mathrm{dv}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}} = u \frac{\mathrm{dv}}{\mathrm{dx}} = F \left(x , v\right)$
but $F \left(x , \frac{\mathrm{dx}}{\mathrm{dt}}\right) = f \left(x\right) \in \left[1 , 9\right]$
$u \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{m} f \left(x\right)$ the force depends on x only not v
u (dv)/dx = 1/m f(x) = 1/m (x^2−x+3)  rearrange and solve by integrating
int vdv = 1/m int(x^2−x+3) dx
${v}^{2} = \frac{2}{5} \left(\frac{1}{3} {x}^{3} - \frac{1}{2} {x}^{2} + 3 x + C\right)$
since the object started with a velocity of 2m/s for $x < 1$
We expect that it was $v = 2 \frac{m}{s}$, for $x = 0$
$C = 10$
$v \left(x\right) = \sqrt{\frac{2}{5} \left(\frac{1}{3} {x}^{3} - \frac{1}{2} {x}^{2} + 3 x + 10\right)}$
$v \left(8\right) = \sqrt{\frac{2}{5} \left(\frac{1}{3} {\left(8\right)}^{3} - \frac{1}{2} {\left(8\right)}^{2} + 38 + 10\right)} = 8.31 \frac{m}{s}$