Problem 64 - Algebra for College Students

Question

The velocity in feet per second (ft'sec) of a rocket \(t\) seconds (sec) after launching is given by the polynomial function \(v(t)=t^{3}-20 t^{2}+110 t .\) What is the velocity of the rocket 10 see after launching? For what value of \(t\) does the rocket have 0 velocity? (GRAPH CANT COPY)

Step-by-Step Solution

Answer

The velocity at 10 seconds is 100 ft/s. The rocket has zero velocity at t = 0.

Step 1: Evaluate the velocity at 10 seconds

To find the velocity of the rocket at 10 seconds, substitute 10 for \(t\) in the given polynomial function: \[ v(10) = 10^3 - 20(10)^2 + 110(10) \] Calculate each term: \[ 10^3 = 1000 \] \[ 20(10)^2 = 2000 \] \[ 110(10) = 1100 \] Now, substitute these values back into the equation: \[ v(10) = 1000 - 2000 + 1100 \] Simplify to find the velocity at 10 seconds: \[ v(10) = 100 \] So, the velocity of the rocket at 10 seconds is 100 ft/s.

Step 2: Set the velocity function to zero to find when the velocity is zero

To find the time \(t\) when the velocity is zero, set the polynomial function equal to zero: \[ t^3 - 20t^2 + 110t = 0 \] Factor out the greatest common factor, which is \(t\): \[ t(t^2 - 20t + 110) = 0 \] This equation is true if either \(t = 0\) or \( t^2 - 20t + 110 = 0 \).

Step 3: Solve the quadratic equation

Solve the quadratic equation \( t^2 - 20t + 110 = 0 \). Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = -20\), and \(c = 110\): \[ t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(110)}}{2(1)} \] This simplifies to: \[ t = \frac{20 \pm \sqrt{400 - 440}}{2} \] \[ t = \frac{20 \pm \sqrt{-40}}{2} \] Since the discriminant (-40) is negative, the quadratic equation has no real roots. Therefore, the only real solution is \(t = 0\).

Key Concepts

Polynomial Functions

A polynomial function is essentially a math formula made from adding, subtracting, and multiplying variables and numbers. Think of it like a recipe that combines different elements to create a unique dish. In our exercise, we have the polynomial function for the velocity of a rocket: \(v(t) = t^3 - 20t^2 + 110t\).
This equation gives us the rocket's velocity at different times after launch. The variable \(t\) stands for time in seconds.

• \(t^3\) represents 't' cubed or 't' multiplied by itself twice.
• \(-20t^2\) means '20 times t squared' with a negative sign in front.
• \(110t\) means '110 times t'.

Notice how each term has a different power of 't'. These powers (3, 2, and 1) tell you that the function is a polynomial and determine its degree, which is 3 (the highest power).

Solving Quadratic Equations

Quadratic equations are a special type of polynomial. They look like this:
\(ax^2 + bx + c = 0 \). In our exercise, when we set the rocket’s velocity to zero, we end up with the quadratic equation \(t^2 - 20t + 110 = 0\).
To solve it, we use the quadratic formula:
\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:

• \(a = 1\)
• \(b = -20\)
• \(c = 110\)

The terms inside the square root (\(b^2 - 4ac\)) have a special name—the discriminant. If the discriminant is negative, like in our exercise (\(-40)\), it means there are no real solutions. So, the only real solution we get when we solve \(t(t^2 - 20t + 110) = 0\) is \(t = 0\).

Factoring

Factoring is like breaking down a big number into smaller building blocks. It's useful in simplifying polynomial functions. In our exercise, we factored out the variable 't' from the equation \(t^3 - 20t^2 + 110t = 0\), giving us:
\(t(t^2 - 20t + 110) = 0\).
This shows that the equation is true if either \(t = 0\) or \(t^2 - 20t + 110 = 0\).
By factoring the equation, we made it easier to isolate 't' and solve for when the rocket’s velocity would be zero. Remember, factoring is a straightforward way to simplify complex problems and make them easier to solve.