Determing Antiderivatives from a Word Problem

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Feb 2008
28
0
"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 2007
666
199
USA
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):

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

There is your final equation.
 
Apr 2005
20,249
7,909
I would have written it slightly differently but basically the same work as Aryth did:

"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?"

One thing you have to be careful about is that speed is give in "miles per hour" while acceleration is given in "feet per second per second". We must change to the same units. There are 5280 feet per mile so 50 miles per hour is 50(5280)= 264000 feet per hour. There are 60 minutes per hour and 60 seconds per minute so there are 3600 seconds per hour. 264000 feet per hour is 26400/3600= 73 and 1/3 feet per second.

Acceleration is the derivative (rate of change) of velocity so \(\displaystyle \frac{dv}{dt}= -22\) (negative because this is deceleration).
Integrating, v= -22t+ C and since it had an initial (t= 0) speed of 73 and 1/3 feet per second, v= -22t+ 73 1/3.

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 \(\displaystyle x= -11t^2+ (73 1/3)t\).