Kevin S. Huang

2014: Problem 10

Topic: Kinematics
Concepts: Angular kinematics, Circular motion

Solution:

Recall the centripetal acceleration is given by

a_c=\omega^2r

Since the car starts at $\omega_0$ and has angular acceleration $\alpha$ ,

\omega=\omega_0+\alpha t

Substituting into the previous equation,

a_c=(\omega_0+\alpha t)^2r

We solve for the time $t$ it takes the car to reach the critical centripetal acceleration $a_c=243\,\mathrm{m/s^2}$ ,

\sqrt{\frac{a_c}{r}}=\omega_0+\alpha t

t=\frac{1}{\alpha}\left(\sqrt{\frac{a_c}{r}}-\omega_0\right)=\frac{1}{4\,\mathrm{s^{-2}}}\left(\sqrt{\frac{243\,\mathrm{m/s^2}}{3\,\mathrm{m}}}-1\,\mathrm{s^{-1}}\right)=2\,\mathrm{s}

so the answer is E.