# How to solve this problem using Calculus?

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? :) – J. M. Sep 7 '11 at 16:29
Clarification to @all: ft/s is the unit feet per sec. I.e., the s is not the same as the distance $s$. – Srivatsan Sep 7 '11 at 16:30
@Ramp: The second derivative denotes acceleration, not velocity; the units of $s''(t)$ would be feet per second square, not feet per second. – Arturo Magidin Sep 7 '11 at 16:31
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] – Mark Bennet Sep 7 '11 at 16:31
@Mark: Wouldn't s(t)=-13/3 t^3 ft/s^2 + cx + d? – Ramp Sep 7 '11 at 16:37

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. – Ramp Sep 7 '11 at 16:47

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. ;) – Ramp Sep 7 '11 at 16:50