A cheetah spots a thomson's gazelle, its preferred prey, and leaps into action, quickly accelerating to its top speed of 30 m/s, the highest of any land animal. however, a cheetah can maintain this extreme speed for only 15 s before having to let up. the cheetah is 170 m from the gazelle as it reaches top speed, and the gazelle sees the cheetah at just this instant. with negligible reaction time, the gazelle heads directly away from the cheetah, accelerating at 4.6 m/s2 for 5.0 s, then running at constant speed. does the gazelle escape? if so, by what distance is the gazelle in front when the cheetah gives up?

We can find the gazelle's constant speed. v = a t = (4.6 m/s^2) (5.0 s) v = 23 m/s Note that the cheetah gains on the gazelle the whole time they are both running. We can find the distance the cheetah can run before it must stop. d = v t = (30 m/s) (15 s) = 450 m To escape, the gazelle has 15 seconds to travel 450 m - 170 m which is 280 meters. We can find the distance x_1 the gazelle travels during the 5.0 second acceleration period. x_1 = (1/2) a t^2 x_1 = (1/2) (4.6 m/s^2) (5.0 s) ^2 x_1 = 57.5 m We can find the distance x_2 the gazelle could run in the next 10 seconds. x_2 = v t = (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 only needed to run 280 meters to escape, the gazelle is able to escape from the cheetah with 7.5 meters to spare.