Question

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`?

Answer 1

`v(x) = sqrt(2/5 (1/3x^3 - 1/2x^2 + 3x +10) )`
`v(8) = sqrt(2/5 (1/3(8)^3 - 1/2(8)^2 + 38 +10) ) = 8.31m/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:
`Fdt = d(mv) = mdv`
So first thing we do it convert this to be in terms of `x`. From above:
`F(x) = m (dv)/dt` re-write in terms of x
`F(x) = m color(red)(d^2x)/dt^2`
we have a differential equation of the form:
`(d^2x)/dt^2 = 1/mF(x, dx/dt)`
Let `v = dx/dt;`
using chain rule `(dv)/dt = (dv)/dx (dx)/dt = u (dv)/dx=F(x, v)`
but `F(x, dx/dt) = f(x) in [1,9]`
`u (dv)/dx = 1/m f(x)` 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 = 2/5 (1/3x^3 - 1/2x^2 + 3x +C)`
since the object started with a velocity of 2m/s for `x<1`
We expect that it was `v = 2m/s`, for `x=0`
`C = 10`
`v(x) = sqrt(2/5 (1/3x^3 - 1/2x^2 + 3x +10) )`
`v(8) = sqrt(2/5 (1/3(8)^3 - 1/2(8)^2 + 38 +10) ) = 8.31m/s`