Question
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
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
Answers
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
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