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)




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.


solving this results in



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.

  • $\begingroup$ 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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