# Find Displacement given force

Find Δx
Given:
Fxe-(t/T)
v0=-41.2 m/s
tf=84.54 s
T=46 s
Fx=13.4 N
m=8.8 kg

F=ma

## The Attempt at a Solution

a=Fxe-(t/T)/m
dv/dt=Fxe-(t/T)/m

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

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

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

cepheid
Staff Emeritus
Gold Member

Find Δx
Given:
Fxe-(t/T)
v0=-41.2 m/s
tf=84.54 s
T=46 s
Fx=13.4 N
m=8.8 kg

F=ma

## The Attempt at a Solution

a=Fxe-(t/T)/m
dv/dt=Fxe-(t/T)/m

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

*take the definite integral between t0 and tf*
x=[(T2)Fx)/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?