Problem 14 - Calculus: Early Transcendentals

Question

Average velocity The position of an object moving vertically along a line is given by the function \(s(t)=-4.9 t^{2}+30 t+20\) Find the average velocity of the object over the following intervals. a. [0,3] b. [0,2] c. [0,1] d. \([0, h],\) where \(h>0\) is a real number

Step-by-Step Solution

Answer

Based on the given position function \(s(t)=-4.9t^2+30t+20\), the average velocities for the specified intervals are: a. The average velocity over the interval [0,3] is approximately 11.7 m/s. b. The average velocity over the interval [0,2] is 20.2 m/s. c. The average velocity over the interval [0,1] is 45.1 m/s. d. The average velocity over the interval [0,h], where h>0, is given by the expression \(-4.9h+30\) m/s.

Step 1: Identify the given intervals and average velocity formula

We are given the position function \(s(t)=-4.9 t^{2}+30 t+20\). The intervals for which we have to find the average velocity are: a. [0,3] b. [0,2] c. [0,1] d. \([0, h],\) where \(h>0\) is a real number Average velocity can be found using the formula: $$\text{Average velocity} = \frac{\Delta s}{\Delta t}$$ where \(\Delta s=s(t_2)-s(t_1)\) and \(\Delta t=t_2-t_1\).

Step 2: Find the average velocity for interval [0,3]

For the interval [0,3], we have \(t_1=0\) and \(t_2=3\). Now calculate the change in position and the change in time: $$\Delta s = s(3)-s(0)$$ $$\Delta t = 3-0$$ Substitute and evaluate to find the change in position: $$\Delta s = (-4.9(3)^2+30(3)+20) - (-4.9(0)^2+30(0)+20) = -4.9(9)+90+20 = 35.1$$ $$\Delta t = 3$$ Now, use the average velocity formula: $$\text{Average velocity} = \frac{35.1}{3} \approx 11.7 \,\text{m/s}$$

Step 3: Find the average velocity for interval [0,2]

For the interval [0,2], we have \(t_1=0\) and \(t_2=2\). Now calculate the change in position and the change in time: $$\Delta s = s(2)-s(0)$$ $$\Delta t = 2-0$$ Substitute and evaluate to find the change in position: $$\Delta s = (-4.9(2)^2+30(2)+20) - (-4.9(0)^2+30(0)+20) = -4.9(4)+60+20 = 40.4$$ $$\Delta t = 2$$ Now, use the average velocity formula: $$\text{Average velocity} = \frac{40.4}{2} = 20.2 \,\text{m/s}$$

Step 4: Find the average velocity for interval [0,1]

For the interval [0,1], we have \(t_1=0\) and \(t_2=1\). Now calculate the change in position and the change in time: $$\Delta s = s(1)-s(0)$$ $$\Delta t = 1-0$$ Substitute and evaluate to find the change in position: $$\Delta s = (-4.9(1)^2+30(1)+20) - (-4.9(0)^2+30(0)+20) = -4.9+30+20 = 45.1$$ $$\Delta t = 1$$ Now, use the average velocity formula: $$\text{Average velocity} = \frac{45.1}{1} = 45.1 \,\text{m/s}$$

Step 5: Find the average velocity for interval [0,h], where h>0

For the interval [0,h], we have \(t_1=0\) and \(t_2=h\). Now calculate the change in position and the change in time: $$\Delta s = s(h)-s(0)$$ $$\Delta t = h-0$$ Substitute and evaluate to find the change in position: $$\Delta s = (-4.9h^2+30h+20) - (-4.9(0)^2+30(0)+20) = -4.9h^2+30h$$ $$\Delta t = h$$ Now, use the average velocity formula: $$\text{Average velocity} = \frac{-4.9h^2+30h}{h}$$ Simplify the expression: $$\text{Average velocity} = -4.9h+30$$ Now, we have the average velocity for all given intervals: a. 11.7 m/s b. 20.2 m/s c. 45.1 m/s d. \(-4.9h+30\) m/s for \(h>0\)

Key Concepts

Position Function

In mathematics, a position function describes the location of an object in space over time. It's a function of time that allows you to compute where an object is at any given moment. A common representation of a position function involves a quadratic equation, especially when describing motion under constant acceleration.

In the given exercise, the position function is given by:

• \( s(t) = -4.9 t^2 + 30t + 20 \)

This equation models vertical motion. Here:

• The term \(-4.9 t^2\) accounts for gravitational acceleration (in meters per second squared, approximately half of the 9.8 m/s² value for gravitational acceleration down)
• The term \(30t\) represents the initial velocity of the object
• The constant term \(20\) represents the initial position of the object

Interval Notation

Interval notation is a mathematical shorthand used to describe an interval, or a range of values. It's imperative to understand interval notation when discussing functions or calculating averages over specific segments.

In the exercise, intervals are provided in the following notations:

• [a, b], where a represents the start and b the end of the interval
• \([0,3], [0,2], [0,1], \text{and} [0,h]\)

Each interval indicates that we calculate the average velocity over this specific range of time \(t\), starting at \(t = 0\) and ending at the specified value (such as \(t = 3\) for the first interval). Understanding this notation helps to focus calculations on the correct sections of the position function.

Mathematical Calculation

In order to compute average velocity, we need to understand the process of calculating the change in position over a change in time. The formula used is:

• \( \text{Average velocity} = \frac{\Delta s}{\Delta t} \)

where:

• \( \Delta s = s(t_2) - s(t_1) \)
• \( \Delta t = t_2 - t_1 \)

To find \( \Delta s \), calculate the position at the end of the interval \( (t = t_2) \) and subtract the position at the start \( (t = t_1) \). Then, divide by the total length of time (\( \Delta t \)).

This approach is repeated for each interval specified in the exercise, including the special interval \([0, h]\), where \(h\) is variable.

Quadratic Equation

A quadratic equation in standard form is represented as \( ax^2 + bx + c = 0 \). Such equations are second-degree polynomials and are foundational in describing parabolic trajectories in physics. They are central to problems involving gravity, like those seen in our exercise's context.

In the position function \( s(t) = -4.9t^2 + 30t + 20 \), the formula fits the pattern of a quadratic equation and provides the height at time \(t\). Understanding the components:

• \(a = -4.9\)
• \(b = 30\)
• \(c = 20\)

The negative coefficient of \(t^2\) implies the object is affected by gravity, giving it a downward, parabolic path. Understanding how these quadratics work allows one to compute key characteristics such as maximum height or the time it takes for an object to reach the ground.