# 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?

## Quadratic Formula:

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!

## Answer and Explanation:

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.

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#### Learn more about this topic:

from Math 101: College Algebra

Chapter 4 / Lesson 10