Determing Antiderivatives from a Word Problem

• Feb 27th 2008, 09:27 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/ $s^{2}$. 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, 10:54 PM
Aryth
Well, we can simply use this one formula:

$a = c$

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

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

$\int -22dx$

$= -22x + C$

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

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

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

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

And we know it slows down to 0, so:

$v_f = 0$

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

We can generalize a formula from this:

$v_f = v_0 + at$

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

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

$\int [v_0 + at]dt$

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

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:

$D = h_0$

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

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

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

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

Acceleration is the derivative (rate of change) of velocity so $\frac{dv}{dt}= -22$ (negative because this is deceleration).
v= dx/dt= -22t+ 73 1/3 so $x= -11t^2+ (73 1/3)t+ C$. 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 $x= -11t^2+ (73 1/3)t$.