Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The tangential acceleration is $1.5~m/s^2$. We can find the speed $v$ when the centripetal acceleration is $1.5~m/s^2$. $a_c = \frac{v^2}{r} = 1.5~m/s^2$ $v = \sqrt{(1.5~m/s^2)(r)}$ $v = \sqrt{(1.5~m/s^2)(100~m)}$ $v = 12.25~m/s$ We can find the time it takes to reach this speed; $t = \frac{v-v_0}{a}$ $t = \frac{12.25~m/s-0}{1.5~m/s^2}$ $t = 8.2~s$ The magnitude of the centripetal acceleration is equal to the tangential acceleration after 8.2 seconds.