Math Problem Statement

The velocity graph represents a vehicle's velocity in feet per minute during a three-minute trip. Answer the following questions: A. What is a method for finding the average velocity during the third minute? B. Determine the units representing the rate of change of the velocity at 3 minutes. C. Interpret v'(3) based on the graph.

Solution

Let's analyze the velocity vs. time graph and address the given questions.


A. What is a method for finding the average velocity during the third minute? (3 points)

To find the average velocity during the third minute (from t=2t = 2 to t=3t = 3):

  1. Identify the velocity values at t=2t = 2 and t=3t = 3 from the graph.
    • From the graph: v(2)2.5feet per minute,v(3)6.5feet per minute.v(2) \approx -2.5 \, \text{feet per minute}, \quad v(3) \approx -6.5 \, \text{feet per minute}.
  2. Use the average value formula: Average velocity=v(2)+v(3)2.\text{Average velocity} = \frac{v(2) + v(3)}{2}. Substituting values: Average velocity=2.5+(6.5)2=92=4.5feet per minute.\text{Average velocity} = \frac{-2.5 + (-6.5)}{2} = \frac{-9}{2} = -4.5 \, \text{feet per minute}.

Answer: The average velocity during the third minute is approximately -4.5 feet per minute.


B. Determine the units representing the rate of change of the velocity at 3 minutes. (3 points)

The rate of change of velocity is the slope of the velocity graph, which is equivalent to acceleration. Its units can be determined as follows:

  1. Velocity is given in feet per minute (ft/min).
  2. Time is measured in minutes (min).
  3. Therefore, the rate of change of velocity (acceleration) is: VelocityTime=ft/minmin=ft/min2.\frac{\text{Velocity}}{\text{Time}} = \frac{\text{ft/min}}{\text{min}} = \text{ft/min}^2.

Answer: The units are feet per minute squared (ft/min²).


C. Interpret v(3)v'(3) based on the graph. (4 points)

The value v(3)v'(3) represents the rate of change of velocity (i.e., acceleration) at t=3t = 3.

To estimate v(3)v'(3) from the graph:

  1. At t=3t = 3, observe the slope of the tangent line to the velocity curve.
  2. From the graph:
    • Between t=2.5t = 2.5 and t=3.5t = 3.5, the velocity drops sharply from approximately 5-5 to 7.5-7.5.
  3. Calculate the slope: v(3)ΔvΔt=7.5(5)3.52.5=7.5+51=2.51=2.5ft/min2.v'(3) \approx \frac{\Delta v}{\Delta t} = \frac{-7.5 - (-5)}{3.5 - 2.5} = \frac{-7.5 + 5}{1} = \frac{-2.5}{1} = -2.5 \, \text{ft/min}^2.

Answer: v(3)2.5ft/min2v'(3) \approx -2.5 \, \text{ft/min}^2. This means the velocity is decreasing at a rate of 2.5ft/min2-2.5 \, \text{ft/min}^2 at t=3t = 3.


Do you have questions or want detailed steps for any part?
Here are 5 related questions to explore:

  1. What is the difference between average velocity and instantaneous velocity?
  2. How would you calculate the total distance traveled during the first 3 minutes?
  3. What does a negative slope on the velocity graph signify?
  4. How can you find the total change in velocity over a time interval?
  5. What physical situations might produce a graph like this one?

Tip: To analyze a velocity-time graph effectively, always note the slopes for acceleration and the area under the curve for displacement.

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Math Problem Analysis

Mathematical Concepts

Velocity
Acceleration
Average rate of change
Instantaneous rate of change

Formulas

Average velocity = (v(t1) + v(t2)) / 2
Rate of change = Δv / Δt

Theorems

Slope of a tangent line for instantaneous rate of change

Suitable Grade Level

Grades 9-12