Question

A car turns a corner at 10mph. Of it were to turn the corner at 30mph, the centripetal acceleration would be

Answer 1

Answer:

9 times larger

Explanation:

The centripetal acceleration of an object in uniform circular motion is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

The car in this problem is moving in a turn, so it is in a circular motion, where r is the radius of the curve. We see that the centripetal acceleration is proportional to the square of the speed, $v^2$.

Let's assume that the initial speed is v = 10 mph, and so the centripetal acceleration is

$a=\frac{v^2}{r}$

Later, the car's speed increases to 30 mph, which is 3 times the original value:

v' = 3v

So, the new centripetal acceleration is

$a'=\frac{v'^2}{r}=\frac{(3v)^2}{r}=9\frac{v^2}{r}=9a$

So, 9 times the original acceleration.