Problem 28 - Calculus - AP Edition

Question

An automobile is driven down a straight highway such that after \(0 \leq t \leq 12\) seconds it is \(s=4.5 t^{2}\) feet from its initial position. (a) Find the average velocity of the car over the interval [0,12]. (b) Find the instantaneous velocity of the car at \(t=6.\)

Step-by-Step Solution

Answer

(a) 54 feet/second, (b) 54 feet/second.

Step 1: Understand the Given Function

The distance function is given by \( s(t) = 4.5t^2 \), representing the distance in feet from the initial position of the car as a function of time \( t \) in seconds.

Step 2: Define Average Velocity Formula

The average velocity over an interval \([a, b]\) is defined as the change in position divided by the change in time: \( v_{avg} = \frac{s(b) - s(a)}{b - a} \). Here, \( a = 0 \) and \( b = 12 \).

Step 3: Calculate Average Velocity (Part a)

First, find \( s(12) = 4.5 \times 12^2 = 648 \) feet. Then, find \( s(0) = 4.5 \times 0^2 = 0 \) feet. The average velocity is \( v_{avg} = \frac{648 - 0}{12 - 0} = 54 \) feet per second.

Step 4: Define Instantaneous Velocity Formula Using Derivatives

Instantaneous velocity is the derivative of the position function \( s(t) \) with respect to time \( t \). So, \( v(t) = \frac{ds}{dt} = \frac{d}{dt}(4.5t^2) \).

Step 5: Calculate Derivative for Instantaneous Velocity

Differentiate \( s(t) = 4.5t^2 \) to get \( v(t) = 9t \) by applying the power rule \( \frac{d}{dt}(t^n) = nt^{n-1} \).

Step 6: Find Instantaneous Velocity at t=6 (Part b)

Substitute \( t = 6 \) into \( v(t) = 9t \) to get \( v(6) = 9 \times 6 = 54 \) feet per second.

Key Concepts

Average Velocity

Average velocity provides a simplified look at how fast something is moving over a period of time. It is the total change in position divided by the total time taken. This gives an overview of motion between two points. In mathematical terms, it's calculated as:\[ v_{avg} = \frac{s(b) - s(a)}{b - a} \]where:

• \( s(b) \) is the position at the end of the time interval.
• \( s(a) \) is the position at the start of the time interval.
• \( b - a \) is the duration of the time interval.

In our example, the car moves from 0 to 648 feet over 12 seconds. This makes the average velocity 54 feet per second. It's like saying on average, the car moved 54 feet forward every second during this time.

Instantaneous Velocity

Instantaneous velocity is a bit different—it shows the speed of an object at a specific moment. Think of it as a snapshot of how fast something is moving right then, without considering other moments. You get this by finding the slope of the tangent line to the position function at that specific point.
In clear mathematical terms, it's the derivative of the position function with respect to time, denoted as \( v(t) \). For the function \( s(t) = 4.5t^2 \), the instantaneous velocity at time \( t = 6 \) has been worked out to be 54 feet per second. This instant velocity tells us exactly how fast the car was going at 6 seconds into its journey.

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a quantity changes. When you differentiate a function, you are essentially finding how one variable changes with respect to another—in this case, distance with respect to time.For a quadratic position function like \( s(t) = 4.5t^2 \), you use derivatives to find instantaneous velocity by applying rules like the power rule, which simplifies this process. The resulting derivative, \( v(t) = 9t \), shows how the car's velocity changes at any given moment in time.

Power Rule

The power rule is one of the simplest and most important techniques in calculus for finding derivatives. This rule states:\[ \frac{d}{dt}(t^n) = nt^{n-1} \]This means you multiply the power \( n \) by the coefficient, and then reduce the power of \( t \) by one. In our example, applying the power rule to \( 4.5t^2 \) gives:

• Multiply 2 (the exponent) by 4.5 (the coefficient), resulting in 9.
• Subtract 1 from the exponent of \( t^2 \) to get \( t^1 \).

This results in the velocity function \( v(t) = 9t \). Understanding the power rule makes it straightforward to differentiate similar polynomial functions and find instantaneous rates of change.