/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A bird is flying due east. Its d... [FREE SOLUTION] | 91Ó°ÊÓ

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A bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{~m}+(12.4 \mathrm{~m} / \mathrm{s}) t-\left(0.0450 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} .\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{~s} ?\)

Short Answer

Expert verified
The instantaneous velocity of the bird at \(t=8.00\,s\) is \(4.24\,m/s\).

Step by step solution

01

Identify the position function

The position of the bird at time \(t\) is given by the function \(x(t) = 28.0\,m +(12.4\,m/s) t-(0.0450\,m/s^3) t^3\).
02

Find the derivative of the position function

We know from calculus that velocity is the derivative of the position. So we need to find the derivative of \(x(t)\) to get the velocity function. Using the power rule, the derivative of \(x(t)\) is \(v(t) = \frac{dx}{dt} = 12.4\,m/s - 3 \times 0.0450\,m/s^3 \times t^2\). Simplifying it gives \(v(t) = 12.4\,m/s - 0.135\,m/s^3 \times t^2\).
03

Substitute the given time into the velocity function

To find the velocity at time \(t=8.00\,s\), we substitute \(t = 8.00\,s\) into the velocity function. This gives \(v(8.00\,s) = 12.4\,m/s - 0.135\,m/s^3 \times (8.00\,s)^2\).
04

Calculate the instantaneous velocity

Carrying out the above calculation yields \(v(8.00\,s) = 12.4 - 0.135 \times 64 = 4.24\,m/s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Position Function
At the heart of understanding motion in physics is the position function, which describes how an object's position changes over time. In the case of our bird flying due east, the position function is given by
\( x(t) = 28.0\text{ m} + (12.4\text{ m/s})t - (0.0450\text{ m/s}^3)t^3 \).
This equation encapsulates the entire motion of the bird with respect to time, accounting for an initial position, constant velocity, and acceleration effects due to, for example, wind resistance or changes in the bird's speed.

The first term, 28.0 m, represents the initial distance of the bird from the building; it's the starting point. The linear term with coefficient 12.4 m/s represents the influence of a constant speed on the position over time. Lastly, the cubic term with coefficient -0.0450 m/s³ illustrates a more complex change in movement, possibly an acceleration component that impacts the bird's speed. Understanding this function is crucial for predicting the bird's location at any given time.
Derivative Calculus
Derivative calculus is a fundamental concept in mathematics that provides us with the tools to describe change. For instance, if we want to know how quickly the bird's position is changing, we refer to its velocity, which is the first derivative of the position function with respect to time.

The process of finding this derivative involves using rules of calculus like the power rule, which states that the derivative of \( t^n \) is \( n \times t^{(n-1)} \). In our example, applying the power rule to each term in the position function \( x(t) \) yields the velocity function:
\( v(t) = \frac{dx}{dt} = 12.4\text{ m/s} - 3 \times 0.0450\text{ m/s}^3 \times t^2 \),
which simplifies to \( v(t) = 12.4\text{ m/s} - 0.135\text{ m/s}^3 \times t^2 \).

By taking the derivative, we've transitioned from describing the bird's location to describing how its speed changes over time, providing insight into its dynamic behaviour.
Velocity Function
The velocity function is a real-time snapshot of the bird's speed for any given moment. It expresses the rate at which the bird's position is changing. After finding the derivative of the position function, we obtained the velocity function:\( v(t) = 12.4\text{ m/s} - 0.135\text{ m/s}^3 \times t^2 \).

To calculate the instantaneous velocity at a specific time, we substitute the desired time into our velocity function. For time \( t=8.00\text{ s} \), we calculate:
\( v(8.00\text{ s}) = 12.4\text{ m/s} - 0.135\text{ m/s}^3 \times (8.00\text{ s})^2 \), which simplifies to \( v(8.00\text{ s}) = 12.4\text{ m/s} - 0.135\text{ m/s}^3 \times 64 = 4.24\text{ m/s} \).

This value is the bird's instantaneous velocity at the eighth second of its flight. It is essential for understanding movements at an exact moment and differs from average velocity, which would consider the total displacement and time elapsed. Real-world applications of this could be determining the exact moment to photograph the bird in flight or predicting its position for collision avoidance systems.

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Most popular questions from this chapter

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of \(3.50 \mathrm{~m} / \mathrm{s}^{2}\) upward. At \(25.0 \mathrm{~s}\) after launch, the second stage fires for \(10.0 \mathrm{~s}\), which boosts the rocket's velocity to \(132.5 \mathrm{~m} / \mathrm{s}\) upward at \(35.0 \mathrm{~s}\) after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stagetwo rocket be moving just as it reaches the launch pad?

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You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for \(10.0 \mathrm{~s}\), Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back \(7.0 \mathrm{~s}\) after leaving the helicopter, and then he has a constant downward acceleration with magnitude \(2.0 \mathrm{~m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

A Honda Civic travels in a straight line along a road. The car's distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t)=\alpha t^{2}-\beta t^{3},\) where \(\alpha=1.50 \mathrm{~m} / \mathrm{s}^{2}\) and \(\beta=0.0500 \mathrm{~m} / \mathrm{s}^{3} .\) Calculate the average velocity of the car for each time interval: (a) \(t=0\) to \(t=2.00 \mathrm{~s} ;\) (b) \(t=0\) to \(t=4.00 \mathrm{~s} ;\) (c) \(t=2.00 \mathrm{~s}\) to \(t=4.00 \mathrm{~s}\)

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