Question

A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s

Answer 1

The centripetal acceleration is 30 m/s²

Step-by-step explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

`\omega=a* t`

Where,

a = acceleration

t = time

Put the value into the formula

`\omega=2.0*5.0`

`\omega=10\ rad/s`

Using formula of centripetal acceleration

`a_(c)=r\omega^2`

`a_(c)=0.30*10^2`

`a_(c)=30\ m/s^2`

Hence, The centripetal acceleration is 30 m/s²