Question

In a roundabout (or traffic circle), cars go around a 27-m-diameter circle. If a car's tires will skid when the car experiences a centripetal acceleration greater than 0.60g, what is the maximum speed of the car in this roundabout?

Answer 1

Answer:

Speed of the car, v = 8.90 m/s

Explanation:

It is given that,

The centripetal acceleration experienced by the car, $a = 0.6g=0.6\times 9.8=5.88\ m/s^2$

Diameter of the circle, d = 27 m

Radius of the circle, r = 13.5 m

The centripetal acceleration is experienced by an object if it moves in a circular path. The formula for the centripetal acceleration is given by :

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

$v=\sqrt{ar}$

$v=\sqrt{5.88\times 13.5}$        

v = 8.90 m/s

So, the maximum speed of the car in this roundabout is 8.9 m/s. Hence, this is the required solution.