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anonymous
 4 years ago
Newport & Vernonville are 200 miles apart.
A car leaves Newport traveling towards Vernonville, and another car leaves Vernonville the same time, traveling towards newport.
The car leaving Newport averages 10mph more than other, and they meet after 1 hour 36. What are the average speeds of the car?
The car leaving Newport averages _______MPH?
The car leaving Vernonville averages _____MPH?
anonymous
 4 years ago
Newport & Vernonville are 200 miles apart. A car leaves Newport traveling towards Vernonville, and another car leaves Vernonville the same time, traveling towards newport. The car leaving Newport averages 10mph more than other, and they meet after 1 hour 36. What are the average speeds of the car? The car leaving Newport averages _______MPH? The car leaving Vernonville averages _____MPH?

This Question is Closed

radar
 4 years ago
Best ResponseYou've already chosen the best response.0The distance that each car travels sums up to 200 miles. distance = rate times time. I am thinkin you might could use that info.

radar
 4 years ago
Best ResponseYou've already chosen the best response.0Let x = distance of the Newport bound car Let y =distance of the veronville bound car x+y=200 We are given time t as 1 hour 36 min, which we need to convert to hour units 1 + 36/60 =1.6 hours. Do I need to go on?

radar
 4 years ago
Best ResponseYou've already chosen the best response.0Mph of Newport bound car = x/1.6 mph of Vernonville bound car = y/1.6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, so my 1st equation is 1.6x+1.6y=200 my second equation is xy10

radar
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think so, lets see what robtobey has to say.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0r t = d Let r be speed of the slower car. Solve the following for r:\[(r+10)\frac{8}{5}+r\frac{8}{5} =200\]\[r=\frac{115}{2} \]\[r+10=\frac{135}{2} \]

radar
 4 years ago
Best ResponseYou've already chosen the best response.0I may have got off to a wrong startt.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\{67.5 \text{ mph},57.5\text{ mph}\} \]

radar
 4 years ago
Best ResponseYou've already chosen the best response.0LynDor please note, 8/5 is the time 1.6 hr

radar
 4 years ago
Best ResponseYou've already chosen the best response.0As you can see I should of Let x =rate of slower car rather than distant, I also should of kept it in terms of one unknown (x+10) rate of faster car.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks. I would never have answered it. So 67.5 is the car leaving newport?

radar
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, the faster car is leaving Newport.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK, thanks. Is there a limit of questions per day?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do I give a medal?

radar
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think there is a limit to the number of questions. To award a medal click on the "Good Answer" button. Good luck with your studies.
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