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Chapter 5: Problem 32

During an Olympic bobsled run, a European team takes a turn of radius \(25 \mathrm{ft}\) at a speed of \(60 \mathrm{mi} / \mathrm{h}\). What acceleration do the riders experience \((a)\) in \(\mathrm{ft} / \mathrm{s}^{2}\) and \((b)\) in units of \(g\) ?

Short Answer

Expert verified
The riders experience an acceleration of \(a = 309.76 ft/s^2\) or equivalently \(9.62 g\).

Step by step solution

01

Convert speed from miles/hour to feet/second

First, convert the speed from miles per hour to feet per second. There are 5280 feet in one mile and 3600 seconds in one hour. So, \(v = 60 mi/hr * (5280 ft / 1 mi) * (1 hr / 3600 s) = 88 ft/s\).
02

Substitute values into the acceleration formula

Next, use the equation \(a = v^2/r\) to find the acceleration. Substitute \(v = 88 ft/s\) and \(r = 25 ft\) into the equation to get \(a = (88 ft/s)^2 / 25 ft = 309.76 ft/s^2\) .
03

Convert acceleration from \(ft/s^2\) to \(g\)

Finally, convert the acceleration from \(ft/s^2\) to \(g\). Since \(1 g = 32.2 ft/s^2\), divide 309.76 by 32.2 to get the acceleration in terms of \(g\). So, \(a_g = 309.76 ft/s^2 / 32.2 ft/s^2/g = 9.62 g\).

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Most popular questions from this chapter

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