## 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?

The distance that each car travels sums up to 200 miles. distance = rate times time. I am thinkin you might could use that info.

Let 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?

Mph of Newport bound car = x/1.6 mph of Vernonville bound car = y/1.6

4. anonymous

Yes, so my 1st equation is 1.6x+1.6y=200 my second equation is x-y-10

I don't think so, lets see what robtobey has to say.

6. anonymous

r 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}$

I may have got off to a wrong startt.

Thanks robtobey.

9. anonymous

$\{67.5 \text{ mph},57.5\text{ mph}\}$

10. anonymous

LynDor please note, 8/5 is the time 1.6 hr

As 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.

13. anonymous

Thanks. I would never have answered it. So 67.5 is the car leaving newport?

Yes, the faster car is leaving Newport.

15. anonymous

OK, thanks. Is there a limit of questions per day?

16. anonymous

How do I give a medal?