an object dropped from the top of a tall building falls y = 16t^2 feet in the first t seconds. Find the average speed during the first 3 seconds of the fall. Find the speed of the object at t = 3 and confirm you answer algebraically. Write out all the written work for these problems

Let's solve the problem step by step.

### Problem 1: Find the average speed during the first 3 seconds of the fall.

We are given the equation for the position (height fallen) as:

\[
y = 16t^2
\]

where \(y\) is the distance fallen in feet, and \(t\) is the time in seconds.

The average speed over a time interval \([0, 3]\) is defined as:

\[
\text{Average speed} = \frac{\text{Total distance fallen}}{\text{Total time}}
\]

First, we calculate the total distance fallen in the first 3 seconds by plugging \(t = 3\) into the equation for \(y\):

\[
y(3) = 16(3)^2 = 16 \times 9 = 144 \text{ feet}
\]

At \(t = 0\), the object hasn't fallen yet, so \(y(0) = 0\).

Thus, the total distance fallen in the first 3 seconds is:

\[
y(3) - y(0) = 144 - 0 = 144 \text{ feet}
\]

The total time is \(3\) seconds.

Now, calculate the average speed:

\[
\text{Average speed} = \frac{144 \text{ feet}}{3 \text{ seconds}} = 48 \text{ feet per second}
\]

### Problem 2: Find the speed of the object at \(t = 3\).

The speed of the object at any time is the derivative of the position function with respect to time, which gives us the instantaneous velocity.

The position function is:

\[
y = 16t^2
\]

To find the velocity, we differentiate \(y\) with respect to \(t\):

\[
v(t) = \frac{dy}{dt} = \frac{d}{dt}(16t^2) = 32t
\]

Now, plug in \(t = 3\) to find the velocity at \(t = 3\):

\[
v(3) = 32(3) = 96 \text{ feet per second}
\]

Thus, the speed of the object at \(t = 3\) seconds is \(96\) feet per second.

### Confirmation of the Answer Algebraically

We can confirm that our average speed answer is consistent with the instantaneous velocity. The average speed during the first 3 seconds was calculated as \(48\) feet per second, while the instantaneous velocity at \(t = 3\) is \(96\) feet per second.

Since the object starts from rest, its velocity increases linearly over time (due to the constant acceleration in free fall), meaning the average speed over the interval should be less than the instantaneous velocity at the end of the interval, which holds true here.

### Summary:
- The average speed during the first 3 seconds is \(48\) feet per second.
- The speed of the object at \(t = 3\) is \(96\) feet per second.

Would you like more details on any step, or do you have any questions?

Here are five related questions you might want to explore:
1. What is the object’s speed at \(t = 1.5\) seconds?
2. How long would it take for the object to fall \(400\) feet?
3. What is the total distance fallen after 5 seconds?
4. How does the object's acceleration affect its velocity over time?
5. What would the equation be if the object were dropped on a planet with different gravity?

**Tip:** The derivative of a position function gives you the velocity function, while the derivative of the velocity function gives you acceleration.

An object dropped from the top of a tall building falls y = 16t^2 feet in the first t seconds. Find the average speed during the first 3 seconds of the fall. Find the speed of the object at t = 3 and confirm your answer algebraically. Write out all the written work for these problems.

Learn how to calculate the average speed and instantaneous velocity of an object dropped from a tall building using the position function y = 16t^2. This detailed solution explores key physics concepts such as kinematics, average speed, and derivatives. Suitable for high school students in Grades 9-12.

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