A c-17 leaves an airport at 5:00 a.m. Heading to a destination 980 miles away.

Algebra
Tutor: None Selected Time limit: 1 Day

It needs to be escorted by an f-18. There is a small area they can meet in. It is from 875-900 miles away. I need to find what time the f18 can leave so that it meet in the area of 875-900. the speed of a c-17 is 518mph and the speed of an f18 is 1190mph. I need to find the earliest time it can leave and the latest time to make that window.

Aug 18th, 2015

Thank you for the opportunity to help you with your question!

We need to determine relative speed. Since they are moving away from each other relative speed

Determining earliest time: earliest time will if they meet at 875 miles

Time taken by c17 =

time = distance/speed = 875/518 =1.69hrs

Time taken by f-18 =

875/1190 =0.74hrs

difference in time = 1.69- 0.74 = 0.95 hrs = 57 minutes

latest time will be if they meet at 900 miles

time by  c17 = 900/518 = 1.74 hrs

time by f18 = 900/1190 =0.76hrs

difference in time = 1.74 -0.76 = 0.98hrs =58.8 minutes =59 minutes

earliest time to leabve will be 5.57 am

lattest time will be 5.59 am

Please let me know if you need any clarification. I'm always happy to answer your questions.
Aug 18th, 2015

Ignore

We need to determine relative speed. Since they are moving away from each other relative speed

Answer should be as below


Aug 18th, 2015

Determining earliest time: earliest time will if they meet at 875 miles

Time taken by c17 =

time = distance/speed = 875/518 =1.69hrs

Time taken by f-18 =

875/1190 =0.74hrs

difference in time = 1.69- 0.74 = 0.95 hrs = 57 minutes

latest time will be if they meet at 900 miles

time by  c17 = 900/518 = 1.74 hrs

time by f18 = 900/1190 =0.76hrs

difference in time = 1.74 -0.76 = 0.98hrs =58.8 minutes =59 minutes

earliest time to leave will be 5.57 am

latest time will be 5.59 am


Aug 18th, 2015

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Aug 18th, 2015
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