/*! 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 183 A truck is traveling along the h... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 183

A truck is traveling along the horizontal circular curve of radius \(r=60 \mathrm{~m}\) with a constant speed \(v=20 \mathrm{~m} / \mathrm{s}\) Determine the angular rate of rotation \(\dot{\theta}\) of the radial line \(r\) and the magnitude of the truck's acceleration.

Short Answer

Expert verified
The angular rate of rotation of the truck is \(\dot{\theta} = 0.3333 rad/sec\) and the magnitude of the truck's acceleration is \(a = 6.6667 m/s^2.\)

Step by step solution

01

Calculating Angular Velocity

First, calculate the angular velocity (\(\dot{\theta}\)) using the formula \(\dot{\theta} = v / r\). Here \(v = 20 m/s\) is the speed, \(r = 60 m\) is the radius of the circle. So, \(\dot{\theta} = 20 / 60 = 0.3333 rad/sec.\) The angular velocity is therefore \(\dot{\theta} = 0.3333 rad/sec.\)
02

Calculating Acceleration

Next, calculate the truck's acceleration. In circular motion, the magnitude of acceleration \(a\) is given by the formula \(a = v^2 / r\), where \(v\) is the speed and \(r\) is the radius of the circle. Using the given values, \(v = 20 m/s\) and \(r = 60 m\), we have \(a = 20^2 / 60 = 6.6667 m/s^2.\) So, the magnitude of the truck's acceleration is \(a = 6.6667 m/s^2.\)

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Most popular questions from this chapter

The position of a crate sliding down a ramp is given by \(x=\left(0.25 t^{3}\right) \mathrm{m}, y=\left(1.5 t^{2}\right) \mathrm{m}, z=\left(6-0.75 t^{5 / 2}\right) \mathrm{m}\), where \(t\) is in seconds. Determine the magnitude of the crate's velocity and acceleration when \(t=2 \mathrm{~s}\)

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