## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the distance the cheetah can run before it must stop. $d = v~t$ $d = (30~m/s)(15~s)$ $d = 450~m$ We can find the distance $x_1$ the gazelle travels during the 5.0 second acceleration period. $x_1 = \frac{1}{2}at^2$ $x_1 = \frac{1}{2}(4.6~m/s^2)(5.0~s)^2$ $x_1 = 57.5~m$ We can find the gazelle's constant speed. $v = a~t$ $v = (4.6~m/s^2)(5.0~s)$ $v = 23~m/s$ We can find the distance $x_2$ the gazelle could run in the next 10 seconds. $x_2 = v~t$ $x_2 = (23~m/s)(10~s)$ $x_2 = 230~m$ The total distance the gazelle can travel in 15 seconds is 230 m + 57.5 m which is 287.5 meters. Since the gazelle had a 170 meter head start, the gazelle is a total of 457.5 meters away from the cheetah's starting point. Since the cheetah must stop after running 450 meters, the gazelle is able to escape.