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

(a) We can equate the distance traveled by David and Tina to find the time it takes Tina to pass David. $\frac{1}{2}at^2=v~t$ $t = \frac{2v}{a} = \frac{(2)(30~m/s)}{2.0~m/s^2}$ $t = 30~s$ We can find the distance that Tina travels in 30 seconds. $x = \frac{1}{2}at^2$ $x = \frac{1}{2}(2.0~m/s^2)(30~s)^2$ $x = 900~m$ Tina drives 900 meters. (b) $v = a~t = (2.0~m/s^2)(30~s)$ $v = 60~m/s$