## Physics: Principles with Applications (7th Edition)

a) $\alpha$ = -3.07012 rad/ $\sec^{2}$ b) t=12.44s c) The car has travelled 95.03 meters before stopping.
a) To find the angular acceleration, the following formula can be used: $\omega_{f}^{2}$ = $\omega_{o}^{2}$ + 2$\alpha$$\theta. This formula is best because we either know or can calculate all the unknown variables and no unnecessary variables, such as time, are present. 1. The given speeds 55km/h and 95km/h are the translational speeds of the wheels. However, by using the formula v=\omegar, they can be converted into angular speeds \omega_{o} and \omega_{f}. \frac{55km}{h}(\frac{1000m}{1km})(\frac{1h}{3600s}) = \frac{275m}{18s} \frac{95km}{h}(\frac{1000m}{1km})(\frac{1h}{3600s}) = \frac{475m}{18s} Translational speed (v) = \omega(radius) (\frac{275m}{18s}) \div (0.4m) = \frac{1375rad}{36s} = \omega_{f} (\frac{475m}{18s}) \div (0.4m) = \frac{2375rad}{36s} = \omega_{o} 2. \theta is the total angle (in radians) through which the wheel has rotated as it slowed down. Since the tires make 75 revolutions and one revolution is 2\pi, the total angle \theta = 75(2\pi) = 150\pi 3. Then we just plug all the variables in the formula \omega_{f}^{2} = \omega_{o}^{2} + 2\alpha$$\theta$ and find the angular acceleration. $\frac{1375rad}{36s}^{2}$ = $\frac{2375rad}{36s}^{2}$ + 2$\alpha$(150$\pi$) $\alpha$ = -3.07012 rad/ $\sec^{2}$ b) The answer can be determined using the formula $\omega_{f}$ = $\omega_{o}$ + $\alpha$t Stopping means that $\omega_{f}$ = 0rad/s and "how much more time" implies that $\omega_{o}$ = $\frac{1375rad}{36s}$. Plug the variables in the formula and solve for t. 0rad/s = $\frac{1375rad}{36s}$ + (-3.07012 rad/$\sec^{2}$)t t=12.44s c) We again use the formula $\omega_{f}^{2}$ = $\omega_{o}^{2}$ + 2$\alpha$$\theta$, only this time to find $\theta$. $\frac{0rad}{s}^{2}$ = $\frac{1375rad}{36s}^{2}$ + 2(-3.07012 rad/$\sec^{2}$)$\theta$ $\theta$ = 237.583 rad The relationship between the distance traveled and the angle theta is d=$\theta$(radius) d = (237.583)(0.4m) = 95.03 m The car has travelled 95.03 meters before stopping