How Do You Calculate Displacement with Varying Force Over Time?

[getty102]

Homework Statement

Find Δx
Given:
F_xe^-(t/T)
v₀=-41.2 m/s
t_f=84.54 s
T=46 s
F_x=13.4 N
m=8.8 kg

Homework Equations

F=ma

The Attempt at a Solution

a=F_xe^-(t/T)/m
dv/dt=F_xe^-(t/T)/m

*integrate both sides*
v=[(-T)F_x)/m]e^-(t/T)+v₀
dx/dt=[(-T)F_x/m]e^-(t/T)+v₀

*take the definite integral between t₀ and t_f*
x=[(T²)F_x)/m]e^-(t/T)+v₀t

I am doing something wrong as my answer of 6192.301 m, is not correct

 

[cepheid]

Homework Statement

Find Δx
Given:
F_xe^-(t/T)
v₀=-41.2 m/s
t_f=84.54 s
T=46 s
F_x=13.4 N
m=8.8 kg

Homework Equations

F=ma

The Attempt at a Solution

a=F_xe^-(t/T)/m
dv/dt=F_xe^-(t/T)/m

*integrate both sides*
v=[(-T)F_x)/m]e^-(t/T)+v₀
dx/dt=[(-T)F_x/m]e^-(t/T)+v₀

*take the definite integral between t₀ and t_f*
x=[(T²)F_x)/m]e^-(t/T)+v₀t

I am doing something wrong as my answer of 6192.301 m, is not correct

I think perhaps your problem comes in assuming that the constant of integration is equal to v₀ (in red above). That is not the case in this situation. To see this, use the generic symbol "C" for the constant of integration here:$$v(t) = \int a(t)\,dt = \frac{F_x}{m}\int e^{-t/T}\,dt = -\frac{T F_x}{m} e^{-t/T} + C$$Now, to solve for C, plug t = 0 into this expression. Note that e⁰ = 1, NOT 0:$$v(0) = v_0 = -\frac{T F_x}{m} + C \\ \Rightarrow C = v_0 + \frac{T F_x}{m}$$If we plug this expression for C back into the expression for v(t), we end up with:$$v(t) = v_0 + \frac{T F_x}{m}(1 - e^{-t/T})$$Can you take it from here?