During a homework problem I was supposed to solve this using Calculus. Here's the problem:
A
cowcar 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 $s''(t) = -26 \ \text{ft}/\text{s}^2$.
Best Answer
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 $$\begin{align*} \text{distance traveled} &= \int_0^{110/39}v(t)\,dt\\ &= \int_0^{110/39}\left(\frac{220}{3} - 26t\right)\,dt \\ &= \frac{220}{3}t - 13t^2\Bigm|_0^{110/39}\\ &= \frac{24200}{117} - \frac{157300}{1521}\\ &\approx 103.4188\text{ feet.} \end{align*}$$