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Suppose an object moves along the y-axis so that its location is y = 2x² +8 at time (where y is measured in meters and is in seconds). (a) Find the average velocity of the object, for a changing from 1 to 5 seconds. (b) Find the instantaneous velocity of the object at 2 seconds.

Suppose an object moves along the y-axis so that its location is y = 2x² +8 at time (where y is measured in meters and is in seconds). (a) Find the average velocity of the object, for a changing from 1 to 5 seconds. (b) Find the instantaneous velocity of the object at 2 seconds.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter1: Expressions And Functions
Section1.8: Interpreting Graphs Of Functions
Problem 1GP
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**Understanding Velocity: Average and Instantaneous** Let's explore how to determine the velocity of an object moving along the y-axis. Given: - The position of the object along the y-axis is defined by the equation: \( y = 2x^2 + 8 \) - Here, \( y \) represents the location in meters and \( x \) is the time in seconds. ### (a) Finding the Average Velocity To find the average velocity of the object over a time interval, we use the formula: \[ \text{Average Velocity} = \frac{\Delta y}{\Delta x} \] where: - \( \Delta y \) is the change in position - \( \Delta x \) is the change in time For \( x \) changing from 1 to 5 seconds: 1. Calculate \( y \) at \( x = 1 \): \[ y(1) = 2(1)^2 + 8 = 10 \] 2. Calculate \( y \) at \( x = 5 \): \[ y(5) = 2(5)^2 + 8 = 58 \] 3. Compute \( \Delta y \) and \( \Delta x \): \[ \Delta y = y(5) - y(1) = 58 - 10 = 48 \] \[ \Delta x = 5 - 1 = 4 \] 4. Find the average velocity: \[ \text{Average Velocity} = \frac{48}{4} = 12 \text{ meters/second} \] ### (b) Finding the Instantaneous Velocity Instantaneous velocity is determined by finding the derivative of the position function with respect to time \( x \). Given the position function: \[ y = 2x^2 + 8 \] The derivative with respect to \( x \) is: \[ \frac{dy}{dx} = 4x \] To find the instantaneous velocity at \( x = 2 \) seconds: \[ \left. \frac{dy}{dx} \right|_{x=2} = 4(2) = 8 \text{ meters/second} \] Thus, the instantaneous velocity of the object at 2 seconds is 8 meters/second.
Transcribed Image Text:**Understanding Velocity: Average and Instantaneous** Let's explore how to determine the velocity of an object moving along the y-axis. Given: - The position of the object along the y-axis is defined by the equation: \( y = 2x^2 + 8 \) - Here, \( y \) represents the location in meters and \( x \) is the time in seconds. ### (a) Finding the Average Velocity To find the average velocity of the object over a time interval, we use the formula: \[ \text{Average Velocity} = \frac{\Delta y}{\Delta x} \] where: - \( \Delta y \) is the change in position - \( \Delta x \) is the change in time For \( x \) changing from 1 to 5 seconds: 1. Calculate \( y \) at \( x = 1 \): \[ y(1) = 2(1)^2 + 8 = 10 \] 2. Calculate \( y \) at \( x = 5 \): \[ y(5) = 2(5)^2 + 8 = 58 \] 3. Compute \( \Delta y \) and \( \Delta x \): \[ \Delta y = y(5) - y(1) = 58 - 10 = 48 \] \[ \Delta x = 5 - 1 = 4 \] 4. Find the average velocity: \[ \text{Average Velocity} = \frac{48}{4} = 12 \text{ meters/second} \] ### (b) Finding the Instantaneous Velocity Instantaneous velocity is determined by finding the derivative of the position function with respect to time \( x \). Given the position function: \[ y = 2x^2 + 8 \] The derivative with respect to \( x \) is: \[ \frac{dy}{dx} = 4x \] To find the instantaneous velocity at \( x = 2 \) seconds: \[ \left. \frac{dy}{dx} \right|_{x=2} = 4(2) = 8 \text{ meters/second} \] Thus, the instantaneous velocity of the object at 2 seconds is 8 meters/second.
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