# Use the given information to write a system of equations, and then solve the system algebraically...

## Question:

Use the given information to write a system of equations, and then solve the system algebraically to answer the question.

A car begins at rest and accelerates. Its distance in meters, t, is modeled by the formula {eq}C(t)= 4t^2 {/eq}. A second car, 150 meters ahead, is traveling at a constant speed of 20 meters per second. Its distance, C(t), in meters can be determined as a function of time, t, in seconds by the formula C(t)= 20t + 150. How long after the first car accelerates will the cars be side by side?

Given a quadratic equation of the form {eq}ax^2 + bx + c = 0 {/eq}, you can always solve for {eq}x {/eq} using the Quadratic Formula. Sometimes factoring may be easier but not all quadratic equations can be easily factor and these are the cases when the Quadratic Formula comes in handy!

The distance of the first car is modeled by the equation

{eq}C(t) = 4t^2 {/eq}

The distance of the second car is modeled

{eq}C(t) = 20t+150 {/eq}

To determine when the cars will be side by side, that is to determine when the distance functions are equal.

{eq}4t^2 = 20t+150 ~~ \implies ~~ 4t^2 -20t -150 = 0 {/eq}

Using the quadratic formula to solve, we get

{eq}t = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-(-20)\pm \sqrt{(-20)^2-4(4)(-150)}}{2(4)} = \frac{20 \pm \sqrt{2800}}{8} {/eq}

This gives

{eq}t = \frac{20 - \sqrt{2800}}{8} = -4.11 {/eq}

or

{eq}t = \frac{20 +\sqrt{2800}}{8} = 9.11 {/eq}

Note that {eq}t {/eq} cannot be negative since it represents time. Therefore the cars will be side by side after 9.11 seconds.