Problem 28 - Calculus Early Transcendentals

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) Average velocity = 54 ft/s, (b) Instantaneous velocity at \(t=6\) is 54 ft/s.

Step 1: Understand the Problem

We are given a position function of a car as a function of time: \( s(t) = 4.5t^2 \). The problem asks for two velocities: the average velocity over the interval \([0, 12]\) and the instantaneous velocity at \(t = 6\).

Step 2: Find the Average Velocity

The average velocity of an object over a time interval \([a, b]\) is given by: \( \text{Average velocity} = \frac{s(b) - s(a)}{b - a} \). Here, \(a = 0\) and \(b = 12\). Calculate \(s(12)\) and \(s(0)\). Substitute these values to find the average velocity.

Step 3: Calculate the Position at Endpoints

Compute \( s(12) = 4.5 \times (12)^2 \) and \( s(0) = 4.5 \times (0)^2 \) to find the position of the car at \(t = 12\) and \(t = 0\).

Step 4: Evaluate and Solve for Velocities

Calculate \(s(12) = 4.5 \times 144 = 648\) and \(s(0) = 0\). Substitute these values into the average velocity formula: \( \text{Average velocity} = \frac{648 - 0}{12 - 0} = 54 \) ft/s.

Step 5: Find the Instantaneous Velocity

The instantaneous velocity is the derivative of the position function with respect to time, \( v(t) = s'(t) \). Calculate this derivative \( v(t) = \frac{d}{dt}(4.5t^2) = 9t \).

Step 6: Evaluate at Specific Time

To find the instantaneous velocity at \(t = 6\), substitute \(t = 6\) into the derivative: \( v(6) = 9 \times 6 = 54 \) ft/s.

Key Concepts

Position Function

A position function describes an object's location over time. In the exercise, the position function is given as \(s(t) = 4.5t^2\), expressing the car's position on a straight highway in feet, relative to its starting point.

• The term \(s(t)\) represents the distance from the initial position as a function of time \(t\), measured in seconds.
• The coefficient \(4.5\) multiplies the squared time \(t^2\), indicating how quickly the car's distance changes as time increases.

To fully understand the change in position, consider that squaring time means the car moves faster with the passage of each second. Thus, the position function \(s(t)\) provides a comprehensive picture of how an object's position evolves continuously over time.

Derivative Calculation

To find the instantaneous velocity of the car, we need to take the derivative of the position function \(s(t)\). Derivatives allow us to understand how a quantity changes at any specific instant. For \(s(t) = 4.5t^2\), the derivative is computed as follows:

• The basic rule of taking the derivative of \(t^n\) results in \(n \times t^{n-1}\).
• Apply this rule: starting with \(s(t) = 4.5t^2\), the derivative \(s'(t)\) is \(9t\).

This derivative, \(s'(t) = 9t\), gives us the car's instantaneous velocity at any time \(t\). Hence, by calculating the derivative, we uncover critical insights into the car's speed dynamics at precise moments during its journey.

Velocity Calculation Steps

Understanding velocity calculation involves determining both average and instantaneous velocities of the car.For average velocity over a time interval, such as \([0, 12]\):

• Use the formula \(\text{Average velocity} = \frac{s(b) - s(a)}{b - a}\).
• Calculate the position at the end points: \(s(0) = 4.5 \times 0^2 = 0\) and \(s(12) = 4.5 \times 144 = 648\).
• Substitute into the formula: \(\text{Average velocity} = \frac{648 - 0}{12 - 0} = 54\) ft/s.

For instantaneous velocity at \(t = 6\):

• Use the derivative \(s'(t) = 9t\).
• Evaluate at \(t = 6\): \(v(6) = 9 \times 6 = 54\) ft/s.

These steps illustrate the process of discerning how velocity changes over an interval and at a specific point in time, connecting the mathematics of calculus to real-world motion.