Question

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`?

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)` the force depends on x only

`u (du)/dx = 1/m f(x) = e^(2x)` rearrange and solve by integrating
`int udu = 1/m inte^(2x) dx`
`1/2u^2 = 1/m e^(2x) + C`
since the object started with a velocity of 2m/s
`C = 4;` and initial velocity, `v_o = sqrt(C) = 2`
`v = sqrt(1/3 e^(2x) ) + 2`

Evaluate velocity at `x = 6; v_(x=6) = sqrt(1/3e^(2*6)) +2 ~~235`
Now Impulse, `I = m (v_2-v_1) ~~ 6(233) = 1398 Kg-m/s`