I think what is meant is that the car's initial velocity is 20 m/s. Let's let the car's position at time \(t=0\) be \(x(0)=0\). The velocity of the car will undergo some kind of exponential decay:
[MATH]v(t)=v_0e^{-kt}[/MATH] where \(0<k\)
Hence:
[MATH]x(t)=\frac{v_0}{k}\left(1-e^{-kt}\right)[/MATH]
Now, we are told:
[MATH]x(t+1)-x(t)=0.92(x(t)-x(t-1))[/MATH]
[MATH]x(t+1)-1.92x(t)+0.92x(t-1)=0[/MATH]
[MATH]\frac{v_0}{k}\left(1-e^{-k(t+1)}\right)-1.92\frac{v_0}{k}\left(1-e^{-kt}\right)+0.92\frac{v_0}{k}\left(1-e^{-k(t-1)}\right)=0[/MATH]
[MATH]\left(1-e^{-k(t+1)}\right)-1.92\left(1-e^{-kt}\right)+0.92\left(1-e^{-k(t-1)}\right)=0[/MATH]
[MATH]e^{-k(t+1)}-1.92e^{-kt}+0.92e^{-k(t-1)}=0[/MATH]
[MATH]0.92e^{2k}-1.92e^{k}+1=0[/MATH]
[MATH](e^{k}-1)(23e^{k}-25)=0[/MATH]
This implies:
[MATH]e^{-k}=\frac{23}{25}[/MATH]
Hence:
[MATH]x(t)=\frac{v_0}{\ln\left(\dfrac{25}{23}\right)}\left(1-\left(\frac{23}{25}\right)^t\right)[/MATH]
Here's a graph of the car's position:
View attachment 12219
And so the total distance \(T\) the car will travel is given by:
[MATH]T=\lim_{t\to\infty}x(t)=\,?[/MATH]