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?

2 Answers
Feb 4, 2016

let's write the impulse equation:
F(x) = m dv/dt re-write in terms of x
F(x) = m (d^2x)/dt^2
we have a differential equation of the form:
(d^2x)/dt^2 = 1/mF(x, dx/dt)
Let u = dx/dt;
using chain rule (du)/dt = (du)/dx (dx)/dt = u (du)/dx
u (du)/dx = 1/mF(x, u) = 1/m f(x)
Note now the Force function is a function of x ONLY.
u (du)/dx = 1/m f(x) = 2x^2 -x rearrange and solve by integrating
int udu = 1/m int(2x^2 - x) dx
1/2u^2 = 1/m (2/3x^3 - 1/2 x^2) + 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(x) = m dv/dt re-write in terms of x
F(x) = m (d^2x)/dt^2
we have a differential equation of the form:
(d^2x)/dt^2 = 1/mF(x, dx/dt)
Let u = dx/dt;
using chain rule (du)/dt = (du)/dx (dx)/dt = u (du)/dx
u (du)/dx = 1/mF(x, u) = 1/m f(x)

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

now calculate v(3) = 8.323 m/s
And the Impulse, I = mDelta v = 5 * 2.323 = 11.62 N