Math Problem Statement

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

Solution

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=16t2y = 16t^2

where yy is the distance fallen in feet, and tt is the time in seconds.

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

Average speed=Total distance fallenTotal time\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=3t = 3 into the equation for yy:

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

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

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

y(3)y(0)=1440=144 feety(3) - y(0) = 144 - 0 = 144 \text{ feet}

The total time is 33 seconds.

Now, calculate the average speed:

Average speed=144 feet3 seconds=48 feet per second\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=3t = 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=16t2y = 16t^2

To find the velocity, we differentiate yy with respect to tt:

v(t)=dydt=ddt(16t2)=32tv(t) = \frac{dy}{dt} = \frac{d}{dt}(16t^2) = 32t

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

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

Thus, the speed of the object at t=3t = 3 seconds is 9696 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 4848 feet per second, while the instantaneous velocity at t=3t = 3 is 9696 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 4848 feet per second.
  • The speed of the object at t=3t = 3 is 9696 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=1.5t = 1.5 seconds?
  2. How long would it take for the object to fall 400400 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.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Average Speed
Instantaneous Velocity
Derivatives

Formulas

Position function: y = 16t^2
Average speed formula: Total distance / Total time
Derivative of position function: v(t) = dy/dt

Theorems

Kinematic equations for uniformly accelerated motion
Power Rule for Differentiation

Suitable Grade Level

Grades 9-12