Time to catch up etc..?

Question: A cheetah can run @ a max speed of 99km/h and a gazelle can run @ max speed of 76.3km/h (a) If both animals are running @ full speed, with the gazelle 97.6m ahead, how long before the cheetah catches its prey? Answer in units s.. (b) The cheetah can maintain it's max speed for only 7.5s. What's the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape? (After 7.5s the speed of cheetah is less than that of the gazelle.) Answer in units m..

Answer: (a) 1: convert velocities from km/hr to m/s (divide by 3600 seconds/hr and multiply by 1000 m/km) Cheetah = 27.5 m/s Gazelle = 21.2 m/s 2: to set up the equation think about it this way: the time the Cheetah has to run going 27.5 m/s must equal the time the Gazelle has to run going 21.2 m/s with a 97.6m head start. So the equation comes out: 27.5 t = 21.2 t + 97.6 3: put all the variable on one side and real numbers on the other by subtracting 21.2t from each side 27.5t - 21.2t = 97.6 ----> 6.3t = 97.6 4: Solve for t by dividing both sides by 6.3 t = 15.5 s (b) To figure this out, determine how far each animal can run in 7.5 seconds - clearly the Cheetah will have travelled further... Then take the difference and give the gazelle at least that much distance head start. In 7.5 seconds the Cheetah will have gone 206.25m and the Gazelle will have run 159m in that same time. 206.25-159 = 47.25 <-- the gazelle needs slightly more distance than this to have been just out of the Cheetah's reach - say a safe distance would be 47.5-48 meters.

Related Questions

Related Items