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Section 2.8 : Applications of Quadratic Equations

2. Two cars start out at the same spot. One car starts to drive north at 40 mph and 3 hours later the second car starts driving to the east at 60 mph. How long after the first car starts driving does it take for the two cars to be 500 miles apart?

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Let’s start out this problem by defining Car A to be the car that drives 40 mph and Car B to be the car that drives 60 mph. Let’s also let t be the time that Car A is driving. From the problem statement we know that Car B starts 3 hours after Car A and so drives for 3 hours less than Car A. This means that \(t - 3\) is the time that Car B is driving.

Show Step 2

Next let’s set up a sketch for this situation.

Show Step 3

Okay. Now we need to get an equation for this situation. The first thing to notice about our sketch is that we have a right triangle! This means we can relate all three lengths using the Pythagorean Theorem (this is one of the reasons to have a sketch – to see these kinds of things).

The Pythagorean Theorem tells us that,

\[{\left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{Car A drives}}\end{array} \right)^2} + {\left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{Car B drives}}\end{array} \right)^2} = {\left( {500} \right)^2} = 250,000\] Show Step 4

Next, we know that we can find the distance of each car using the formula,

\[{\mbox{Distance}} = \left( {{\mbox{Speed of Car}}} \right)\left( {{\mbox{Time driving}}} \right)\]

So, for each car we have,

\[\begin{array}{l}{\mbox{Distance of Car A}} = \left( {40} \right)\left( t \right) = 40t\\ {\mbox{Distance of Car B}} = \left( {60} \right)\left( {t - 3} \right) = 60\left( {t - 3} \right)\end{array}\]

Putting all of this into the “word equation” we wrote down in Step 3 we get the following equation.

\[\begin{array}{rl}{\left( {40t} \right)^2} + {\left( {60\left( {t - 3} \right)} \right)^2} & = 250,000\\{40^2}{t^2} + {60^2}{\left( {t - 3} \right)^2} & = 250,000\\ 1600{t^2} + 3600\left( {{t^2} - 6t + 9} \right) & = 250,000\\ 1600{t^2} + 3600{t^2} - 21,600t + 32,400 & = 250,000\\5200{t^2} - 21,600t - 217,600 & = 0\end{array}\]

Note as well that we did quite a bit of simplification to get the equation into a standard form. Also, do not get excited about the “large” numbers here! They happen on occasion so they are nothing to worry about. This is still just a quadratic and we know how to solve quadratic equations. It doesn’t matter if the numbers are single digit numbers of significantly larger numbers as they are here.

Show Step 5

As noted in the previous step this is just a quadratic equation and we know how to solve those! Using the quadratic formula gives,

\[t = \frac{{ - \left( { - 21,600} \right) \pm \sqrt {{{\left( { - 21,600} \right)}^2} - 4\left( {5200} \right)\left( { - 217,600} \right)} }}{{2\left( {5200} \right)}} = \frac{{21,600 \pm \sqrt {4,992,640,000} }}{{10,400}}\] Show Step 6

Reducing the two values we got in the previous steps to decimals we arrive at the following two solutions to the quadratic equation from Step 4.

\[t = \frac{{21,600 - \sqrt {4,992,640,000} }}{{10,400}} = - 4.7172\hspace{0.25in}t = \frac{{21,600 + \sqrt {4,992,640,000} }}{{10,400}} = 8.8710\]

The first solution to the equation doesn’t make any sense since it is negative (we are working with time and so it’s safe to assume we are starting at \(t = 0\) after all!) so that means the second is the answer we need.

This means that Car A (i.e. the one traveling at 40 mph) travels for 8.871 hours while Car B (i.e. the one traveling at 60 mph) travels for 5.871 hours (three hours less than Car A time!).