/*! 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

You throw a rock straight up and find that it returns to your hand \(3.60 \mathrm{~s}\) after it left your hand. Neglect air resistance. What was th maximum height above your hand that the rock reached?

A 15 kg rock is dropped from rest on the earth and reaches the ground in \(1.75 \mathrm{~s}\). When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in \(18.6 \mathrm{~s}\). What is the acceleration due to gravity on Enceladus?

A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot \(0.380 \mathrm{~s}\) to pass from the top to the bottom of this window, which is \(1.90 \mathrm{~m}\) high. How far is the top of the window below the windowsill from which the flowerpot fell?

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of \(2.10 \mathrm{~m} / \mathrm{s}^{2},\) and the car has an acceleration of \(3.40 \mathrm{~m} / \mathrm{s}^{2}\). The car overtakes the truck after the truck has moved \(60.0 \mathrm{~m}\). (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x=0\) at the initial location of the truck.

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than \(1.00 \mathrm{~s}\) in the air (their "hang time"). Treat the athlete as a particle and let \(y_{\max }\) be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above \(y_{\max } / 2\) to the time it takes him to go from the floor to that height. Ignore air resistance.

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