# Determing Antiderivatives from a Word Problem

• Feb 27th 2008, 08:27 PM
!!!
Determing Antiderivatives from a Word Problem
"A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/$\displaystyle s^{2}$. What is the distance covered before the car comes to a stop?"

I know that the antiderivative of the acceleration is the velocity, and the antiderivative of the velocity is its distance, but I need help on how to derive the equations from this word problem.
• Feb 27th 2008, 09:54 PM
Aryth
Well, we can simply use this one formula:

$\displaystyle a = c$

Meaning the acceleration is constant, in this case, it is $\displaystyle -22 \frac{ft}{s^2}$

So, we simply take the antiderivative to find the velocity function:

$\displaystyle \int -22dx$

$\displaystyle = -22x + C$

Well, we know that this equals the velocity, but it's in $\displaystyle \frac{mi}{h}$ and not $\displaystyle \frac{ft}{s}$, so we have to convert it.

$\displaystyle 50 \frac{mi}{h} = \frac{220}{3} \frac{ft}{s}$

Now we set the constant equal to the initial velocity, seeing as it fits the equation to find the final equation:

$\displaystyle v_0 = \frac{220}{3} = C$

And we know it slows down to 0, so:

$\displaystyle v_f = 0$

$\displaystyle 0 = -22x + \frac{220}{3}$

We can generalize a formula from this:

$\displaystyle v_f = v_0 + at$

$\displaystyle -22t = at$
$\displaystyle C = v_0$
$\displaystyle 0 = v_f$

Now we take the antiderivative of the right side of that function:

$\displaystyle \int [v_0 + at]dt$

$\displaystyle = v_0t + \frac{1}{2}at^2 + D$

This is the position function, and we can assume that since the initial velocity was the C value, that the initial position value would be the D value:

$\displaystyle D = h_0$

So, this function is a function of position, which is commonly called s(t):

$\displaystyle s(t) = h_0 + v_0t + \frac{1}{2}at^2$

Now we plug in what we know (Assuming we started at position = 0):

$\displaystyle s(t) = \frac{220}{3}t - 11t^2$

Acceleration is the derivative (rate of change) of velocity so $\displaystyle \frac{dv}{dt}= -22$ (negative because this is deceleration).
v= dx/dt= -22t+ 73 1/3 so $\displaystyle x= -11t^2+ (73 1/3)t+ C$. Since we want to know how far the car travels before it stops, we can take x= 0 as the initial position and so C= 0. The car will have stopped when v= 0 so solve -22t+ 73 1/3= 0 for t and put that value of t into $\displaystyle x= -11t^2+ (73 1/3)t$.