Problem 98 - Precalculus Student Solutions Manual 5th

Question

Velocity of an Object The velocity of an object, \(v,\) after \(t\) seconds is given by $$ v=3 t^{2}-18 t+24 $$ Find the interval where the velocity is negative.

Step-by-Step Solution

Answer

The velocity is negative in the interval \( (2, 4) \).

Step 1 - Write down the velocity function

The given velocity function is \[ v(t) = 3t^2 - 18t + 24 \]

Step 2 - Find the roots of the equation

Set the velocity function equal to zero to find the roots. Solve \[ 3t^2 - 18t + 24 = 0 \] Divide the entire equation by 3:\[ t^2 - 6t + 8 = 0 \]Now factorize the quadratic equation:\[ (t-2)(t-4) = 0 \]So the roots are \[ t=2 \] and \[ t=4 \]

Step 3 - Determine the sign of the velocity in each interval

Test the intervals determined by the roots: \[ (-\infty, 2) \ (2, 4) \ (4, +\infty) \]Choose a test point in each interval and substitute it into the velocity function:- For the interval \[ (-\infty, 2) \], choose \( t = 1 \):\[ v(1) = 3(1)^2 - 18(1) + 24 = 3 - 18 + 24 = 9 \] so \( v(1)>0 \) - For the interval \[ (2, 4) \], choose \( t = 3 \):\[ v(3) = 3(3)^2 - 18(3) + 24 = 27 - 54 + 24 = -3 \] so \( v(3)<0 \)- For the interval \[ (4, +\infty) \], choose \( t = 5 \):\[ v(5) = 3(5)^2 - 18(5) + 24 = 75 - 90 + 24 = 9 \] so \( v(5)>0 \)

Step 4 - Identify the interval with negative velocity

From the testing, the velocity is negative in the interval \[ (2, 4) \]

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2. It usually has the form ax^2 + bx + c = 0
It features a term where the variable (usually denoted as x or, in this case, t) is squared. This gives the equation its characteristic parabolic shape when graphed.
To solve a quadratic equation, you can use different methods such as:

• Factoring
• Quadratic formula
• Completing the square

In this exercise, the velocity function is given by ( v(t)=3t^2−18t+24 ). To find the points where it changes sign, you set the quadratic equation to zero and solve for t.
In our example, we had: ( 3t^2−18t+24=0 ). Simplifying by dividing the entire equation by 3, we get ( t^2−6t+8=0 ). From here, we used factoring to find the roots.

Roots of an Equation

The roots of an equation are the solutions for which the equation equals zero. For quadratic equations, these points are crucial. They tell us where the graph intersects the x-axis. To find the roots:

• Step 1: Set the quadratic equation equal to zero: ( t^2 - 6t + 8 = 0 )
• Step 2: Factorize the equation:
  ( (t - 2)(t - 4) = 0 ). This tells us that the roots are t = 2 and t = 4

The roots divide the timeline into intervals where the sign of the velocity might be different:

• ( (−∞, 2)
• ( (2, 4)
• ( (4, +∞)

Testing values within these intervals allows us to determine where the velocity changes from positive to negative or vice versa.

Sign of Velocity

The sign of the velocity function tells us whether the object is moving forward or backward within a given time interval. To determine the sign of the velocity, you substitute values within the divided intervals into the velocity function and check if the result is positive or negative.
Let's review how we did this:

• For ( (-∞, 2) ), substitute t = 1 : ( v(1) = 3(1)^2 - 18(1) + 24 = 9 ). The velocity is positive.
• For ( (2, 4) ), substitute t = 3 : ( v(3) = 3(3)^2 - 18(3) + 24 = -3 ). The velocity is negative.
• For ( (4, +∞) ), substitute t = 5 : ( v(5) = 3(5)^2 - 18(5) + 24 = 9 ). The velocity is positive.

By analyzing these calculations, we can conclude that the velocity of the object is negative in the interval ( (2, 4) ).