Thread: 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

First start with F(t) = ma(t), or rearrange to get .

Next . You can evalaute the constant of integration C from the initial condition that V(0) = 4
m/s.

Next: and again you can find the value for the constant of integration D from the initial condition y(0) = 2 m.

Hope this helps.