SOLUTION: Plese help me out on this.I dont know what equation to use. Here is the problem-A cheeta is 300 ft. from its prey.It starts to sprint towards its prey at 90 ft. per second. At the

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Plese help me out on this.I dont know what equation to use. Here is the problem-A cheeta is 300 ft. from its prey.It starts to sprint towards its prey at 90 ft. per second. At the       Log On


   

Question 56574: Plese help me out on this.I dont know what equation to use. Here is the problem-A cheeta is 300 ft. from its prey.It starts to sprint towards its prey at 90 ft. per second. At the same time, thee prey starts to sprint at 70 ft. per second. When will the cheeta catch its prey?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A cheeta is 300 ft. from its prey.It starts to sprint towards its prey at 90 ft. per second. At the same time, thee prey starts to sprint at 70 ft. per second. When will the cheeta catch its prey?
Let d = distance the prey travel when it is caught
Then (d + 300) = distance the cheetah travels when it catches
:
Both animals will be traveling for same number of seconds
Write time equation: t = dist/speed
:
Cheetah's time = Prey's time
%28d%2B300%29%2F90 = d%2F70
:
Cross mult:
70(d+300) = 90d
70d + 21000 = 90d
21000 = 90d - 70d
20d = 21000
d = 21000/20
d = 1050 ft traveled by the unfortunate one
:
and 1350 ft traveled by the hungry one:
:
Time = 1350/90 = 15 sec
:
Check: 1050/70 = 15 sec also