108km/hr = 108,000m/3600s= 30m/s
75.9km/hr = 75900m/3600s = 21.1m/s
the simplest way to do this is to recognize that the cheetah travels 8.9m/s faster than the gazelle
the cheetah has to make up a difference of 70.9 m with its speed advantage of 8.9m/s, so it will take the cheetah
time = dist/speed = 70.9m/8.9m/s = 8s
if the cheetah can maintain this speed advantage for only 7.5s, it can only close a distance of 8.9m/s x 7.5 s = 66.8m, so this is the minimum safety distance for the gazelle
The cheetah can maintain its maximum speed
for only 7.5 s.
What is the minimum distance the gazelle
must be ahead of the cheetah to have a chance
of escape? (After 7.5 s the speed of cheetah is
less than that of the gazelle.)
Answer in units of m
1 answer