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Let's solve the problem step by step using the given position function \( s(t) = -16t^2 + 101t \).

### 1. Average Velocity Formula:
The average velocity over an interval \([a, b]\) is given by the formula:
\[
v_{\text{avg}} = \frac{s(b) - s(a)}{b - a}
\]
where \(s(t)\) is the position function.

### 2. Calculate the values for each time interval:

#### (a) Time Interval \([2, 3]\):
- \( s(2) = -16(2)^2 + 101(2) = -64 + 202 = 138 \)
- \( s(3) = -16(3)^2 + 101(3) = -144 + 303 = 159 \)

Now, the average velocity:
\[
v_{\text{avg}} = \frac{159 - 138}{3 - 2} = \frac{21}{1} = 21
\]

#### (b) Time Interval \([2.9, 3]\):
- \( s(2.9) = -16(2.9)^2 + 101(2.9) = -16(8.41) + 292.9 = -134.56 + 292.9 = 158.34 \)

Now, the average velocity:
\[
v_{\text{avg}} = \frac{159 - 158.34}{3 - 2.9} = \frac{0.66}{0.1} = 6.6
\]

#### (c) Time Interval \([2.99, 3]\):
- \( s(2.99) = -16(2.99)^2 + 101(2.99) = -16(8.9401) + 302.99 = -143.0416 + 302.99 = 159.9484 \)

Now, the average velocity:
\[
v_{\text{avg}} = \frac{159 - 159.9484}{3 - 2.99} = \frac{0.0516}{0.01} = 5.16
\]

#### (d) Time Interval \([2.999, 3]\):
- \( s(2.999) = -16(2.999)^2 + 101(2.999) = -16(8.994001) + 302.999 = -143.904016 + 302.999 = 159.094984 \)

Now, the average velocity:
\[
v_{\text{avg}} = \frac{159 - 159.094984}{3 - 2.999} = \frac{0.005016}{0.001} = 5.016
\]

#### (e) Time Interval \([2.9999, 3]\):
- \( s(2.9999) = -16(2.9999)^2 + 101(2.9999) = -16(8.99940001) + 302.9999 = -143.99040016 + 302.9999 = 159.00949984 \)

Now, the average velocity:
\[
v_{\text{avg}} = \frac{159 - 159.00949984}{3 - 2.9999} = \frac{0.000500016}{0.0001} = 5.0016
\]

### 3. Instantaneous Velocity at \( t = 3 \):
To find the instantaneous velocity at \( t = 3 \), we take the derivative of \( s(t) \):
\[
s'(t) = \frac{d}{dt}(-16t^2 + 101t) = -32t + 101
\]
Now, plug in \( t = 3 \):
\[
s'(3) = -32(3) + 101 = -96 + 101 = 5
\]

So, the **instantaneous velocity at \( t = 3 \)** is **5**.

Would you like more details or have any further questions? Here are some related questions to expand on this:

1. What is the difference between average and instantaneous velocity?
2. How does taking the derivative of a position function give us velocity?
3. How can we approximate the instantaneous velocity using smaller and smaller time intervals?
4. How do we interpret negative values for velocity?
5. What would the position graph of this function look like?

**Tip:** Instantaneous velocity is just the slope of the tangent line to the position-time curve at a given time!

For the position function s(t) = -16t^2 + 101t, complete the table with appropriate average velocities. Make a conjecture about the value of the instantaneous velocity at t = 3.

Learn how to compute average velocities for different time intervals and estimate the instantaneous velocity at t = 3 using the function s(t) = -16t^2 + 101t. This detailed explanation covers calculus concepts such as derivatives and limits, making it suitable for high school or early college students.

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