Problem 47 - Beginning and Intermediate Algebra

Question

Solve each problem. When appropriate, round answers to the nearest tenth. The following function gives the distance in feet a car going approximately 68 mph will skid in \(t\) seconds. $$ D(t)=13 t^{2}-100 t $$ Find the time it would take for the car to skid \(180 \mathrm{ft}\).

Step-by-Step Solution

Answer

The car will skid for approximately 9.2 seconds.

Step 1: Set Up the Equation

Given the function for distance, we need to find the time it takes for the car to skid 180 feet. Set the function equal to 180:\[ 180 = 13t^2 - 100t \]

Step 2: Rearrange the Equation

Rearrange the equation to set it to zero by subtracting 180 from both sides:\[ 13t^2 - 100t - 180 = 0 \]

Step 3: Apply the Quadratic Formula

Use the quadratic formula to solve for t: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 13 \), \( b = -100 \), and \( c = -180 \). Substitute these values into the formula.\[ t = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 13 \cdot (-180)}}{2 \cdot 13} \]

Step 4: Simplify Under the Square Root

Simplify the expression inside the square root:\[ t = \frac{100 \pm \sqrt{10000 + 9360}}{26} \]

Step 5: Complete the Square Root Calculation

Calculate the value inside the square root and then the entire numerator:\[ t = \frac{100 \pm \sqrt{19360}}{26} \]

Step 6: Solve for t

Calculate \( \sqrt{19360} \). Approximate and simplify:\[ t = \frac{100 \pm 139.1}{26} \] This gives two potential solutions for t. Calculate them individually:\[ t = \frac{100 + 139.1}{26} \approx 9.2 \] \[ t = \frac{100 - 139.1}{26} \approx -1.5 \]

Step 7: Select the Valid Solution

Time cannot be negative, so discard \( t = -1.5 \). The valid solution is:\[ t \approx 9.2 \] seconds

Key Concepts

Distance Function

In this exercise, we are dealing with a distance function, which describes how far a car going 68 mph will skid over time. The function given is: \[ D(t) = 13t^2 - 100t \] Did you notice the equation uses time (\( t \)) as a variable to calculate distance (\( D(t) \))? The coefficients and constants here are based on physical constants and measurements. In simple terms:

• \( t \): Represents time in seconds.
• \( D(t) \): Represents distance in feet.

We need to determine at what point in time the car skids 180 feet. This means we set the equation to 180 and solve for \( t \). Understanding this setup is crucial in moving forward with solving the equation.

Quadratic Formula

Next up is the Quadratic Formula. When we transformed the problem to find \( t \) by setting the equation to equal 180, we ended up with a quadratic equation: \[ 13t^2 - 100t - 180 = 0 \] This type of equation can be solved using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Let's identify what the components \( a \), \( b \), and \( c \) represent in our equation:

• \( a = 13 \)
• \( b = -100 \)
• \( c = -180 \)

Substituting these values into the quadratic formula allows us to simplify the equation further and find the value(s) of \( t \) which are the potential times it would take for the car to skid 180 feet.

Roots of a Quadratic

Finally, let's talk about solving for the roots of a quadratic equation. A quadratic equation can have two solutions, which are known as roots. In our example, after simplifying everything, we get: \[ t = \frac{100 \pm \sqrt{19360}}{26} \] Simplifying further: \[ t = \frac{100 \pm 139.1}{26} \] gives us two roots (potential times):

• \( t \approx 9.2 \)
• \( t \approx -1.5 \)

Roots can be positive or negative, but in our context, time cannot be negative. Therefore, we discard \( t \approx -1.5 \) and keep \( t \approx 9.2 \) as our valid solution. The car will skid approximately 9.2 seconds to cover 180 feet.