Math Problem Statement

Assume that the function s(t) = t^3 - 2t^2 represents the position of an object at time t moving along a line, where s is measured in feet and t is measured in seconds. The exercise asks to calculate the average velocity of the object over various time intervals.

Solution

The exercise asks us to calculate the average velocity of an object over various time intervals using the function s(t)=t32t2s(t) = t^3 - 2t^2, which represents the position of the object at time tt. The average velocity is calculated as the change in position over the change in time for each interval.

The average velocity formula is:

vavg=s(t2)s(t1)t2t1v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

where [t1,t2][t_1, t_2] is the given time interval. Let’s calculate the average velocities for each interval.

  1. For the interval [1,1.5][1, 1.5]:

    • t1=1t_1 = 1, t2=1.5t_2 = 1.5
    • Calculate s(1)s(1) and s(1.5)s(1.5): s(1)=132×12=12=1s(1) = 1^3 - 2 \times 1^2 = 1 - 2 = -1 s(1.5)=(1.5)32×(1.5)2=3.3754.5=1.125s(1.5) = (1.5)^3 - 2 \times (1.5)^2 = 3.375 - 4.5 = -1.125
    • Average velocity: vavg=1.125(1)1.51=0.1250.5=0.25ft/sv_{\text{avg}} = \frac{-1.125 - (-1)}{1.5 - 1} = \frac{-0.125}{0.5} = -0.25 \, \text{ft/s}
  2. For the interval [1,1.1][1, 1.1]:

    • t1=1t_1 = 1, t2=1.1t_2 = 1.1
    • Calculate s(1)s(1) and s(1.1)s(1.1): s(1.1)=(1.1)32×(1.1)2=1.3312.42=1.089s(1.1) = (1.1)^3 - 2 \times (1.1)^2 = 1.331 - 2.42 = -1.089
    • Average velocity: vavg=1.089(1)1.11=0.0890.1=0.89ft/sv_{\text{avg}} = \frac{-1.089 - (-1)}{1.1 - 1} = \frac{-0.089}{0.1} = -0.89 \, \text{ft/s}
  3. For the interval [0.5,1][0.5, 1]:

    • t1=0.5t_1 = 0.5, t2=1t_2 = 1
    • Calculate s(0.5)s(0.5) and s(1)s(1): s(0.5)=(0.5)32×(0.5)2=0.1250.5=0.375s(0.5) = (0.5)^3 - 2 \times (0.5)^2 = 0.125 - 0.5 = -0.375
    • Average velocity: vavg=1(0.375)10.5=0.6250.5=1.25ft/sv_{\text{avg}} = \frac{-1 - (-0.375)}{1 - 0.5} = \frac{-0.625}{0.5} = -1.25 \, \text{ft/s}
  4. For the interval [0.9,1][0.9, 1]:

    • t1=0.9t_1 = 0.9, t2=1t_2 = 1
    • Calculate s(0.9)s(0.9) and s(1)s(1): s(0.9)=(0.9)32×(0.9)2=0.7291.62=0.891s(0.9) = (0.9)^3 - 2 \times (0.9)^2 = 0.729 - 1.62 = -0.891
    • Average velocity: vavg=1(0.891)10.9=0.1090.1=1.09ft/sv_{\text{avg}} = \frac{-1 - (-0.891)}{1 - 0.9} = \frac{-0.109}{0.1} = -1.09 \, \text{ft/s}

Based on these calculations, the reasonable conjecture for the instantaneous velocity at t=1t = 1 is around 1ft/s-1 \, \text{ft/s}.

Would you like further details or explanations?

Related Questions:

  1. How do you find the instantaneous velocity from a function?
  2. What is the difference between average and instantaneous velocity?
  3. How does the derivative relate to velocity?
  4. Can the average velocity be used to estimate instantaneous velocity?
  5. What is the physical interpretation of negative velocity?

Tip: To find the instantaneous velocity at a specific point, calculate the derivative of the position function and evaluate it at that point.

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Math Problem Analysis

Mathematical Concepts

Calculus
Average Velocity
Position Function
Instantaneous Velocity

Formulas

v_avg = (s(t2) - s(t1)) / (t2 - t1)

Theorems

Derivative and Velocity Relationship

Suitable Grade Level

Grades 11-12