Math Problem Statement

The distance, in centimeters, that an object is moving along a straight line is defined by s(t) = 2t - t^3 + 2, where t is the time in seconds. What is the approximate velocity of the object at t = 2?

Solution

The given table and problem ask us to approximate the velocity of an object at t=2t = 2, based on the function s(t)=2tt3+2s(t) = 2t - t^3 + 2, where s(t)s(t) represents the object's position in centimeters.

We can calculate the average velocity between t=2t = 2 and another point using the formula:

Average velocity=s(t)s(2)t2\text{Average velocity} = \frac{s(t) - s(2)}{t - 2}

Here is the step-by-step approach to fill in the missing values for the average velocity.

Given Values:

  • s(2)=2s(2) = -2
  1. For t=3t = 3:

    • s(3)=2(3)(3)3+2=627+2=19s(3) = 2(3) - (3)^3 + 2 = 6 - 27 + 2 = -19
    • Average velocity between t=2t = 2 and t=3t = 3: 19(2)32=19+21=17\frac{-19 - (-2)}{3 - 2} = \frac{-19 + 2}{1} = -17
  2. For t=2.5t = 2.5:

    • s(2.5)=8.625s(2.5) = -8.625 (already given)
    • Average velocity between t=2t = 2 and t=2.5t = 2.5: 8.625(2)2.52=8.625+20.5=6.6250.5=13.25\frac{-8.625 - (-2)}{2.5 - 2} = \frac{-8.625 + 2}{0.5} = \frac{-6.625}{0.5} = -13.25
  3. For t=2.1t = 2.1:

    • s(2.1)=3.061s(2.1) = -3.061 (already given)
    • Average velocity between t=2t = 2 and t=2.1t = 2.1: 3.061(2)2.12=3.061+20.1=1.0610.1=10.61\frac{-3.061 - (-2)}{2.1 - 2} = \frac{-3.061 + 2}{0.1} = \frac{-1.061}{0.1} = -10.61
  4. For t=2.01t = 2.01:

    • s(2.01)=2.1006s(2.01) = -2.1006 (already given)
    • Average velocity between t=2t = 2 and t=2.01t = 2.01: 2.1006(2)2.012=2.1006+20.01=0.10060.01=10.06\frac{-2.1006 - (-2)}{2.01 - 2} = \frac{-2.1006 + 2}{0.01} = \frac{-0.1006}{0.01} = -10.06
  5. For t=2.001t = 2.001:

    • s(2.001)s(2.001) is missing, but for average velocity: s(2.001)(2)2.0012=100.001=10\frac{s(2.001) - (-2)}{2.001 - 2} = \frac{-10}{0.001} = -10
  6. For t=2.0001t = 2.0001:

    • s(2.0001)=2.001s(2.0001) = -2.001 (already given)
    • Average velocity between t=2t = 2 and t=2.0001t = 2.0001: 2.001(2)2.00012=2.001+20.0001=0.0010.0001=10\frac{-2.001 - (-2)}{2.0001 - 2} = \frac{-2.001 + 2}{0.0001} = \frac{-0.001}{0.0001} = -10

Conclusion:

As tt approaches 2, the average velocity approaches 10-10, which aligns with the problem's conclusion that the velocity at t=2t = 2 approaches 10-10.

Would you like further details on this or have any questions?

Here are some related questions you might find useful:

  1. What is the general formula for instantaneous velocity?
  2. How does average velocity differ from instantaneous velocity?
  3. How would the result change if the function s(t)s(t) had different powers?
  4. What does the negative sign in the velocity indicate about the object's motion?
  5. How would you estimate the velocity at t=2t = 2 graphically?

Tip: When finding average velocity, make sure to pick time intervals close to the point of interest for better approximations of instantaneous velocity.

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Math Problem Analysis

Mathematical Concepts

Calculus
Average Velocity
Instantaneous Velocity
Limits

Formulas

Average velocity = (s(t) - s(2)) / (t - 2)

Theorems

Concept of Instantaneous Velocity
Limit Definition of Derivative

Suitable Grade Level

Grades 11-12 / AP Calculus