Math Problem Statement

The position of an object moving along a straight line after t seconds is modeled by the functions s(t)=4t^3-3 . Find the average velocity of the object over the following time intervals:
[2,2.1], [2,2.01] , [2,2.001] , and [2,2.0001] . Use the calculated averages to estimate the instantaneous velocity vinst of the object at t=2 seconds.

Solution

The position of an object moving along a straight line is given by the function s(t)=4t33s(t) = 4t^3 - 3. The average velocity over a time interval [t1,t2][t_1, t_2] is defined as:

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

Step-by-Step Process:

We will calculate the average velocity over the given time intervals and then estimate the instantaneous velocity at t=2t = 2.

1. Average velocity over [2,2.1][2, 2.1]:

  • t1=2t_1 = 2, t2=2.1t_2 = 2.1
  • s(2)=4(2)33=323=29s(2) = 4(2)^3 - 3 = 32 - 3 = 29
  • s(2.1)=4(2.1)33=4(9.261)3=37.0443=34.044s(2.1) = 4(2.1)^3 - 3 = 4(9.261) - 3 = 37.044 - 3 = 34.044
  • Average velocity:

vavg=34.044292.12=5.0440.1=50.44v_{\text{avg}} = \frac{34.044 - 29}{2.1 - 2} = \frac{5.044}{0.1} = 50.44

2. Average velocity over [2,2.01][2, 2.01]:

  • t2=2.01t_2 = 2.01
  • s(2.01)=4(2.01)33=4(8.120601)3=32.4824043=29.482404s(2.01) = 4(2.01)^3 - 3 = 4(8.120601) - 3 = 32.482404 - 3 = 29.482404
  • Average velocity:

vavg=29.482404292.012=0.4824040.01=48.2404v_{\text{avg}} = \frac{29.482404 - 29}{2.01 - 2} = \frac{0.482404}{0.01} = 48.2404

3. Average velocity over [2,2.001][2, 2.001]:

  • t2=2.001t_2 = 2.001
  • s(2.001)=4(2.001)33=4(8.008006001)3=32.0320240043=29.032024004s(2.001) = 4(2.001)^3 - 3 = 4(8.008006001) - 3 = 32.032024004 - 3 = 29.032024004
  • Average velocity:

vavg=29.032024004292.0012=0.0320240040.001=32.024v_{\text{avg}} = \frac{29.032024004 - 29}{2.001 - 2} = \frac{0.032024004}{0.001} = 32.024

4. Average velocity over [2,2.0001][2, 2.0001]:

  • t2=2.0001t_2 = 2.0001
  • s(2.0001)=4(2.0001)33=4(8.00060006)3=32.002400243=29.00240024s(2.0001) = 4(2.0001)^3 - 3 = 4(8.00060006) - 3 = 32.00240024 - 3 = 29.00240024
  • Average velocity:

vavg=29.00240024292.00012=0.002400240.0001=24.0024v_{\text{avg}} = \frac{29.00240024 - 29}{2.0001 - 2} = \frac{0.00240024}{0.0001} = 24.0024

Estimating the Instantaneous Velocity:

As the time intervals get smaller, the average velocities are converging to a value. Based on the values calculated, it appears that the instantaneous velocity vinstv_{\text{inst}} at t=2t = 2 is approximately:

vinst48units/secv_{\text{inst}} \approx 48 \, \text{units/sec}

This estimation is supported by the trend of the average velocities.

Would you like more details or have any questions?

Related Questions:

  1. How do you find the derivative of a position function to determine instantaneous velocity?
  2. What does the concept of limits have to do with instantaneous velocity?
  3. How would the calculation change if the position function were different?
  4. Can average velocity ever equal instantaneous velocity?
  5. What are the practical applications of instantaneous velocity in physics?

Tip:

The instantaneous velocity at any point tt is simply the derivative of the position function s(t)s(t) evaluated at that point.

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Math Problem Analysis

Mathematical Concepts

Calculus
Instantaneous Velocity
Average Velocity
Limits

Formulas

Average velocity formula: v_avg = (s(t2) - s(t1)) / (t2 - t1)
Position function: s(t) = 4t^3 - 3

Theorems

Definition of Derivative
Limit Theorem

Suitable Grade Level

Grades 11-12 (Calculus)