Question

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`?

Answer 1

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)

Answer 2

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`