# An object with a mass of 5 kg is traveling at 6 m/s. If the object is accelerated by a force of f(x) = 2x^2-x over x in [0, 9], where x is in meters, what is the impulse at x = 3?

Feb 4, 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 $\frac{\mathrm{du}}{\mathrm{dt}} = \frac{\mathrm{du}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}} = u \frac{\mathrm{du}}{\mathrm{dx}}$
$u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{m} F \left(x , u\right) = \frac{1}{m} f \left(x\right)$
Note now the Force function is a function of x ONLY.
$u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{m} f \left(x\right) = 2 {x}^{2} - x$ rearrange and solve by integrating
$\int u \mathrm{du} = \frac{1}{m} \int \left(2 {x}^{2} - x\right) \mathrm{dx}$
$\frac{1}{2} {u}^{2} = \frac{1}{m} \left(\frac{2}{3} {x}^{3} - \frac{1}{2} {x}^{2}\right) + C$
The velocity after replacing u by v is:
v = sqrt(2/5 (2/3x^3 - 1/2 x^2) + C
evaluate C by setting v(0) = 6
v(0) = 6 = sqrt(C); C = 36
now calculate v(3)

Feb 4, 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 $\frac{\mathrm{du}}{\mathrm{dt}} = \frac{\mathrm{du}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}} = u \frac{\mathrm{du}}{\mathrm{dx}}$
$u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{m} F \left(x , u\right) = \frac{1}{m} f \left(x\right)$

Note now the Force function is a function of x ONLY.
$u \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{m} f \left(x\right) = 2 {x}^{2} - x$ rearrange and solve by integrating
$\int u \mathrm{du} = \frac{1}{m} \int \left(2 {x}^{2} - x\right) \mathrm{dx}$
$\frac{1}{2} {u}^{2} = \frac{1}{m} \left(\frac{2}{3} {x}^{3} - \frac{1}{2} {x}^{2}\right) + C$
since the object started with a velocity of 6m/s
C = 36; and initial velocity, $v \left(0\right) = \sqrt{C} = 6$
$v = \sqrt{\frac{2}{5} \left(\frac{2}{3} {x}^{3} - \frac{1}{2} {x}^{2}\right)} + 6$

now calculate v(3) = 8.323 m/s
And the Impulse, $I = m \Delta v = 5 \cdot 2.323 = 11.62 N$