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An object moving at constant speed $v$ around a circle of radius $r$ has an acceleration a directed toward the center of the circle. The SI unit of acceleration is $m / s^{2}$ .
If the speed is increased $21.5 %$ , by what percentage does the radial acceleration increase?
$%$

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Answer

So the acceleration increases by $47.2225$ percent.

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Step 1 ：Let the original speed be $v$ and the original radius be $r$ . The original acceleration is $a = \frac{v^{2}}{r}$ .

Step 2 ：The new speed is $\frac{121.5}{100} v$ , so the new acceleration is $a^{'} = \frac{(\frac{121.5}{100} v)^{2}}{r} = {(\frac{121.5}{100})}^{2} \frac{v^{2}}{r}$ .

Step 3 ：This is ${(\frac{121.5}{100})}^{2}$ times the original acceleration, which is an increase of ${(\frac{121.5}{100})}^{2} - 1 = \frac{14722.25}{10000} - 1 = \frac{4722.25}{10000}$ .

Step 4 ：So the acceleration increases by $47.2225$ percent.

Required informationAn object moving at constant speed v around a circle of radius r has an acceleration a directed toward the center of the circle. The SI unit of acceleration is m/s2.If the speed is increased 21.5%, by what percentage does the radial acceleration increase?%

Answer

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Required information
An object moving at constant speed $v$ around a circle of radius $r$ has an acceleration a directed toward the center of the circle. The SI unit of acceleration is $m / s^{2}$ .
If the speed is increased $21.5 %$ , by what percentage does the radial acceleration increase?
$%$