The position of a rock dropped from a height of 100-meters is given by the function s(t)=−5t2+100. Find the average velocity of the rock over the intervals [2.999,3.0] and [3.0,3.001]. Use this information to approximate the instantaneous velocity of the rock at time t=3.0.

(Round your answer to the nearest integer if necessary.)

Provide your answer below:

The position function of the rock is given as \( s(t) = -5t^2 + 100 \). To find the average velocity over the intervals \([2.999, 3.0]\) and \([3.0, 3.001]\), we use the formula for average velocity:

\[
\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

### Interval \([2.999, 3.0]\):
- \( t_1 = 2.999 \), \( t_2 = 3.0 \)
- \( s(2.999) = -5(2.999)^2 + 100 \)
- \( s(3.0) = -5(3.0)^2 + 100 \)

Let's calculate:

\[
s(2.999) = -5(2.999)^2 + 100 = -5(8.994001) + 100 = -44.970005 + 100 = 55.029995
\]
\[
s(3.0) = -5(9.0) + 100 = -45 + 100 = 55
\]
Now, the average velocity for this interval is:

\[
\text{Average velocity} = \frac{55 - 55.029995}{3.0 - 2.999} = \frac{-0.029995}{0.001} = -29.995 \approx -30
\]

### Interval \([3.0, 3.001]\):
- \( t_1 = 3.0 \), \( t_2 = 3.001 \)
- \( s(3.001) = -5(3.001)^2 + 100 \)
- \( s(3.0) = -5(3.0)^2 + 100 = 55 \)

\[
s(3.001) = -5(3.001)^2 + 100 = -5(9.006001) + 100 = -45.030005 + 100 = 54.969995
\]
Now, the average velocity for this interval is:

\[
\text{Average velocity} = \frac{54.969995 - 55}{3.001 - 3.0} = \frac{-0.030005}{0.001} = -30.005 \approx -30
\]

### Instantaneous velocity at \( t = 3.0 \):
The instantaneous velocity is approximated by the average velocity over these small intervals. Since both intervals give approximately \(-30\), we conclude that the instantaneous velocity at \( t = 3.0 \) is approximately:

\[
\boxed{-30 \text{ m/s}}
\]

Would you like more details or have any further questions? Here are some related questions for practice:

1. What is the velocity of the rock at \( t = 2.0 \)?
2. Find the position of the rock at \( t = 4.0 \).
3. What is the total distance traveled by the rock in the first 3 seconds?
4. How do you interpret the negative velocity in terms of the rock's motion?
5. At what time does the rock hit the ground?

**Tip:** Instantaneous velocity can be found as the derivative of the position function \( s(t) \), which represents the rate of change of position at any specific time \( t \).

The position of a rock dropped from a height of 100-meters is given by the function s(t) = -5t^2 + 100. Find the average velocity of the rock over the intervals [2.999, 3.0] and [3.0, 3.001]. Use this information to approximate the instantaneous velocity of the rock at time t = 3.0. (Round your answer to the nearest integer if necessary.)

Calculate the average velocity of a rock dropped from a 100-meter height over small time intervals and approximate its instantaneous velocity at t = 3.0. Learn how to use calculus concepts, such as limits and derivatives, to solve this physics-based problem. Suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation