# An object with a mass of 6 kg is traveling at 2 m/s. If the object is accelerated by a force of f(x) = e^(2x)  over x in [0, 9], where x is in meters, what is the impulse at x = 6?

Feb 7, 2016

let's write the impulse equation:
$F \left(x\right) = m \frac{\mathrm{dv}}{\mathrm{dt}}$ re-write in terms of x
$F \left(x\right) = m \frac{{d}^{2} x}{\mathrm{dt}} ^ 2$
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 u = dx/dt;
using chain rule $\frac{\mathrm{du}}{\mathrm{dt}} = \frac{\mathrm{du}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}} = u \frac{\mathrm{du}}{\mathrm{dx}}$
$u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{m} F \left(x , u\right) = \frac{1}{m} f \left(x\right)$ the force depends on x only

$u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{m} f \left(x\right) = {e}^{2 x}$ rearrange and solve by integrating
$\int u \mathrm{du} = \frac{1}{m} \int {e}^{2 x} \mathrm{dx}$
$\frac{1}{2} {u}^{2} = \frac{1}{m} {e}^{2 x} + C$
since the object started with a velocity of 2m/s
C = 4; and initial velocity, ${v}_{o} = \sqrt{C} = 2$
$v = \sqrt{\frac{1}{3} {e}^{2 x}} + 2$

Evaluate velocity at x = 6; v_(x=6) = sqrt(1/3e^(2*6)) +2 ~~235
Now Impulse, $I = m \left({v}_{2} - {v}_{1}\right) \approx 6 \left(233\right) = 1398 K g - \frac{m}{s}$