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Chapter 12: Problem 110

An automobile is traveling on a curve having a radius of \(800 \mathrm{ft}\). If the acceleration of the automobile is \(5 \mathrm{ft} / \mathrm{s}^{2},\) determine the constant speed at which the automobile is traveling.

Short Answer

Expert verified
The automobile is traveling at a constant speed of approximately 89.44 feet/second.

Step by step solution

01

Understand the problem

We are given that the automobile is traveling on a curve with a radius of 800 feet and has an acceleration of 5 feet per second squared. The goal is to find the constant speed of the automobile.
02

Use the formulas of circular motion

In uniform circular motion, the centripetal (radial) acceleration can be described by the formula \(a = v^{2}/r\), where \(a\) is the acceleration, \(v\) is the speed, and \(r\) is the radius of the circular path. Here, we know \(a\) and \(r\) and we need to solve for \(v\).
03

Solve for speed

Rearrange the formula to solve for speed: \(v = \sqrt{a \cdot r}\). Plug in the values: \(v = \sqrt{5 ft/s^{2} \cdot 800 ft}\). Simply calculate the expression to find the speed.

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Key Concepts

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

Uniform Circular Motion
When an object moves in a perfect circle at a constant speed, it is undergoing what is known as uniform circular motion. Imagine a car rounding a racetrack at a steady pace; the car is continually changing direction, curving along the track, which means it is always accelerating, even if its speedometer reads a constant value.

Key to understanding uniform circular motion is that the speed is uniform鈥攎eaning the magnitude of the velocity stays consistent鈥攂ut because the direction of the velocity changes, the object is still accelerating. This continuous change in velocity's direction is the result of an inward force, known as centripetal force, which keeps the object traveling along the circular path.
Centripetal Acceleration
The continuous acceleration that occurs in uniform circular motion is referred to as centripetal acceleration. It is always directed towards the center of the circle and is a result of the centripetal force.

Centripetal acceleration (\(a\textsubscript{c}\)) keeps an object on its circular path, and its magnitude can be calculated with the formula: \( a = \frac{v^2}{r} \), where \( v \) represents the speed of the object, and \( r \) stands for the radius of the circular path. The centripetal acceleration is essential in understanding circular motion dynamics because it describes how quickly the object's velocity is changing direction.
Circular Motion Speed Calculation
Calculating the speed of an object in uniform circular motion requires understanding the relationship between speed, radius, and centripetal acceleration. By rearranging the centripetal acceleration formula, we can solve for speed as \( v = \sqrt{a \cdot r} \).

To calculate the speed of our automobile from the exercise, we substitute the given values into the rearranged formula: \( v = \sqrt{5 \: ft/s^2 \cdot 800 \: ft} \). This would allow us to solve for the speed at which the automobile must travel to maintain the given centripetal acceleration. Understanding this calculation method is crucial for problems involving circular motion since it provides a direct way to compute the speed from known values of acceleration and radius.

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Most popular questions from this chapter

A sphere is fired downwards into a medium with an initial speed of \(27 \mathrm{m} / \mathrm{s}\). If it experiences a deceleration of \(a=(-6 t) \mathrm{m} / \mathrm{s}^{2},\) where \(t\) is in seconds, determine the distance traveled before it stops.

A man can swim at \(4 \mathrm{ft} / \mathrm{s}\) in still water. He wishes to cross the 40 -ft-wide river to point \(B, 30\) ft downstream. If the river flows with a velocity of \(2 \mathrm{ft} / \mathrm{s}\), determine the speed of the man and the time needed to make the crossing. Note: While in the water he must not direct himself toward point \(B\) to reach this point. Why?

The double collar \(C\) is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod \(A B .\) If the angular velocity of \(A B\) is given as \(\dot{\theta}=\left(e^{0.5 t^{2}}\right)\) rad \(/ \mathrm{s},\) where \(t\) is in seconds, and the path defined by the fixed rod is \(r=|(0.4 \sin \theta+0.2)| \mathrm{m},\) determine the radial and transverse components of the collar's velocity and acceleration when \(t=1\) s. When \(t=0, \theta=0 .\) Use Simpson's rule with \(n=50\) to determine \(\theta\) at \(t=1\) s.

The arm of the robot moves so that \(r=3 \mathrm{ft}\) is constant, and its grip \(A\) moves along the path \(z=(3 \sin 4 \theta) \mathrm{ft}\) where \(\theta\) is in radians. If \(\theta=(0.5 t)\) rad, where \(t\) is in seconds, determine the magnitudes of the grip's velocity and acceleration when \(t=3 \mathrm{s}\).

If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation \(a=9.81\left[1-v^{2}\left(10^{-4}\right)\right] \mathrm{m} / \mathrm{s}^{2}\) where \(v\) is in \(\mathrm{m} / \mathrm{s}\) and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when \(t=5 \mathrm{s},\) and (b) the body's terminal or maximum attainable velocity \((\text { as } t \rightarrow \infty)\).
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