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During a homework problem I was supposed to solve this using Calculus. Here's the problem:

A cow car is traveling at $50 \ \text{mi}/\text{h}$ when the brakes are fully applied, producing a constant deceleration of $26 \ \text{ft}/\text{s}^2$. What is the distance traveled before the car comes to a stop?

So, I just solved this with Physics equations and came up with $103.4 \ \text{ft}$. However, I guess I'm supposed to use Calculus, seeing this is a Calculus class. ;) Any ideas how to do this?

I guess that $s''(t) = -26 \ \text{ft}/\text{s}^2$.

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    Cows have brakes? :)2011-09-07
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    Clarification to @all: `ft/s` is the unit feet per sec. I.e., the `s` is not the same as the distance $s$.2011-09-07
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    @Ramp: The second derivative denotes acceleration, not velocity; the units of $s''(t)$ would be feet per second square, not feet per second.2011-09-07
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    Can you show that applying your calculus knowledge to the equation you have quoted gives you the physics equation you have used to solve the problem [integrate twice and be careful with constants]2011-09-07
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    @Mark: Wouldn't `s(t)=-13/3 t^3 ft/s^2 + cx + d`?2011-09-07
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    @Sriva: Ah, so it would be `s(t)=-13/3 t^3 ft/s^2 + ct + d`. Also, apologies - I don't have a guide to the formatting language used here at mse.2011-09-07
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    `s'(t)=-13t^2 ft/s^2 + c`2011-09-07
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    If I differentiate that, I get $s''(t) = -13 \times 2t + 0$ ignoring units. That is not the one we started with...2011-09-07
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    Ramp, $\LaTeX$ is being used here for math expressions. You can use [this](http://www.codecogs.com/latex/eqneditor.php) to help you with making them.2011-09-07
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    Awesome, thanks JM. Sriv, I am getting the same answer. Take a look in Latex and see if it helps you see what I wrote: $\[s'(t)=-13t^2 ft/s^2 + c\]$2011-09-07
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    Feet, miles, seconds, hours ... has your teacher ever heard of coherent units?2011-09-07
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    @Ramp: You need to use `$` for in-line formulas and `$$` for displayed formula. The backslash is an escape character here, so if you type `\[` it interprets it as "please print a left square bracket", not as "please start the $\LaTeX$ display math mode".2011-09-07
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    @Henning: apparently unit conversion will be a required skill for this course... but rather distracting for the goal of appreciating calculus, no?2011-09-07

2 Answers 2

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Let $v(t)$ be the velocity. You are told that $v(0)=50$ mi/h; you are also told that when brakes are applied, acceleration, $v'(t)$, is $-26\ \mathrm{ft}/\mathrm{sec}^2$. It's better to put everything in the same units, so let's change velocity to feet per second: $$50\frac{\mathrm{mi}}{\mathrm{hr}} = 50\left(\frac{5280\text{ feet}}{1\text{ mi}}\right)\left(\frac{1\text{ hr}}{3600\text{ sec}}\right) = \frac{220}{3}\text{ft}/\text{sec}.$$

So $v(0) = \frac{220}{3}$ ft/sec. $v'(t) = -26$ ft/sec${}^2$.

By the First Fundamental Theorem of Calculus, we have: $$v(t) -v(0) = \int_0^t v'(x)\,dx = \int_0^t(-26)\,dx = -26x\Bigm|_0^t = -26t.$$ Therefore, $v(t) = v(0) - 26t = \frac{220}{3}-26t$.

The car comes to a stop when $v(t)=0$. Solving $v(t)=0$ we get $$t = \frac{220}{78} = \frac{110}{39}\text{ seconds.}$$ How far did it travel? The distance traveled is the integral of the velocity, so $$\begin{align*} \text{distance traveled} &= \int_0^{110/39}v(t)\,dt\\ &= \int_0^{110/39}\left(\frac{220}{3} - 26t\right)\,dt \\ &= \frac{220}{3}t - 13t^2\Bigm|_0^{110/39}\\ &= \frac{24200}{117} - \frac{157300}{1521}\\ &\approx 103.4188\text{ feet.} \end{align*}$$

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    Oh snap I haven't gotten to the fundamental theorem of Calculus yet. Thanks for your help, though.2011-09-07
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For constant acceleration, calculus only justifies the usual equation $s=x_0+v_0t+\frac{1}{2}at^2$. With $a=-26, v_0=\frac{220}{3}, x_0=0$ you get $s=\frac{v_0^2}{2|a|}$. I agree with your answer.

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    Thanks for agreeing. But I need to know how to solve it with Calculus. ;)2011-09-07