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The Question is as follows

A Car is traveling at $88ft/sec$ (60 mph) when the driver applies the brakes to avoid hitting a child. After $t$ seconds, the car is $s(t)=88t-8t^2$ feet from the point where the brakes were first applied. How long does it take for the car to come to a stop, and how far does it travel before stopping?

I've always had problems with related rates and optimization, but I think I have an answer for this although I didn't use calculus to get to my answer, so I'd like some input both on if what I did is correct and if there is a better way to solve the problem.

I figured that I could determine how long the car takes to stop simply by finding the roots of the original equation simply by solving $0=88t-8t^2$

From this I determined that it took the car $11/8$ seconds to stop.

Now that I'm writing this and thinking more I don't believe what I did next is correct. I simply multiplied by $88 ft/sec$ to result in $121 ft$ before the car stopped. I don't think this accounts for the car slowing as it is stopping though, it just assumes a constant speed until the time ends.

I can't really think of what I can do to find the distance using calculus. I'm given the $88ft/sec$ which I believe would be the $\frac{dx}{dt}$, but I'm not sure how to use that properly.

Edit: Alrighty, so s'(t)=v(t)

$s'(t)=88-8t$

$0=88-8t$

$t=\frac{11}{2}$

So it takes 11/2 seconds for the car to stop, and that can be plugged back into the original equation to solve for distance.

$s(\frac{11}{2})=88(\frac{11}{2})-8(\frac{11}{2})^2$

solving this results in

$s(\frac{11}{2})=242$ft

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Why to do think that you will get the time to come to a stop by setting the distance travelled, s(t), to zero? And, how do you get 11/8 s?

First compare the given equation for s(t) with the standard form:

s(t) = u0 * t + a*t^2/2

where u0 is the initial velocity and a the acceleration. We see that u0 = 88 ft/s and a = -16 ft/s^2.

To find the time it takes the car to go from 88 ft/s to zero velocity, use the standard equation

v(t) = u0 + a * t

As the final speed is zero, we get

0 = 88 - 16 * t, where t is now the time taken for the car to come to a stop. This gives t = 11/2 s as the time it takes for the car to stop.

The get the distance traveled, you substitute this time into the original equation that you were given to get 242 ft.

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  • $\begingroup$ Apologies for my idiocy. I felt confident initially, but not at all now. I wasn't sure what s(t) was and thought of it initially as similar to the rock thrown problem, so I could solve to find the roots because the first root at 0 would be the point where the car began breaking and the last would be the point where the car stopped, though I can see thats wrong now. I also was not aware of the standard equations. I see now that taking the derivative of s(t) results s'(t)=88-16t and solving for a critical value and I can plug that value of 11/2 back into the original and get 242ft $\endgroup$ – Vin May 14 '17 at 21:00

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