solving motion of type m x'' = F(t)

A particle of mass 3 kg, initially at position y(0) = 2 and moving with velocity v(0) = 4) is subject to the exponential driving force F(t) = 4e^(-3t)

Find an expression for y(t).

I've got to the point where i've applied v(t) = V(0) + 1/M x (integral) f(t').

no clue what to do after.

any help would be appreciated :)

Re: solving motion of type m x'' = F(t)

First start with F(t) = ma(t), or rearrange to get $\displaystyle a(t) = \frac {F(t)}{m} = \frac 4 3 e^{-3t} $.

Next $\displaystyle v = \int a(t) dt = \int \frac 4 3 e^{-3t} dt = - \frac 4 9 e^{-3t} + C$. You can evalaute the constant of integration C from the initial condition that V(0) = 4

m/s.

Next: $\displaystyle y(t) = \int v(t)dt = \int (-\frac 4 9 e^{-3t} + C) dt = \frac 4 {27} e^{-3t} + C t + D$ and again you can find the value for the constant of integration D from the initial condition y(0) = 2 m.

Hope this helps.