PARTIAL SOLUTION:
Let $y$ represent the distance from the car to the wall and $t$ be the time elapsed. Then you have $$\frac{dy}{dt}=-y$$ $$\frac{dt}{dy}=-\frac{1}{y}$$ $$t=-\int\frac{dy}{y}$$ $$t=-\ln y+C$$ $$y=e^{-t+C}$$ Since the starting distance is $100$ feet, then you can substitute $100$ for $y$ and $0$ for $t$ to find $C$: $$100=e^{C}$$ $$C=\ln 100$$ and so $$y=100e^{-t}$$ Can you use this to solve for the amount of time after which the distance was $10$ feet?