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A car is travelling towards a wall at a rate equal to its distance from the wall.

How long until the car hits the wall?

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closed as off-topic by Leucippus, Zain Patel, Arnaldo, Daniel W. Farlow, Namaste Jun 6 '17 at 23:52

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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?

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  • $\begingroup$ So to find the time it takes for the distance from the wall, or y, I have to plug them into the equation? 10=100e^-t t=ln(10) t=2.3 $\endgroup$ – Jack Murphy Jun 6 '17 at 17:11
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    $\begingroup$ Yep, that's correct. Looks like you got it. :) $\endgroup$ – Frpzzd Jun 6 '17 at 17:13
  • $\begingroup$ Thanks for the help! $\endgroup$ – Jack Murphy Jun 6 '17 at 17:25
  • $\begingroup$ No problem! If you found it helpful, don't forget to $\uparrow$! :) $\endgroup$ – Frpzzd Jun 6 '17 at 17:33
  • $\begingroup$ Where did you get $100$ feet as the starting distance? I don't see it in the question. $\endgroup$ – Ross Millikan Jun 6 '17 at 21:10

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