Distance Problem

D

debmcgary

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Joined
Jul 16, 2006
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I don't know why I can't wrap my head around the distance problems. I have tried and tried to figure out an easy way to understand.

If I have 2 vehicles that leave for a locatin that is 280 miles away and the rate of 1 vehicle is twice that of the other, and the faster vehicle arrives 4 hours earlier than the other, how do I figure the rate of the slower vehicle.

Any help would be appreciated.

thanks

deb
 
Okay, letting x be speed of slower car and t be time it would take car to go 280 miles, then xt = 280, and 2x(t-4) = 280, or 2xt - 8x = 280.
Can you solve it from there?
 
No, I just have a mental block when it comes to this. I have asked my teacher and he is very patient, but I am just not getting it.
 
Okay, since xt = 280, then 2xt = 2*280 = 560, right!
And since 2xt - 8x =280, then 560 - 8x = 280, or -8x = -280.
So, x = -280/-8 = 35.
 
Thanks for your help. It still doesn't make sense to me though. I know you were solving for x. But (I know I really sound stupid), in the equation, what happened to t?

Sorry if I am too thick to get this whole process, but it has been several years since I have been out of school and I am trying going back after 30 years!!

Yikes!!
 
Okay, i was able to substitute 2xt for 560, becuase since xt= 280, then 2xt = 560, right?
So, then 2xt - 8x = 280 becomes 560 - 8x = 280.
 
"Okay, letting x be speed of slower car and t be time it would take car to go 280 miles, then xt = 280, and 2x(t-4) = 280, or 2xt - 8x = 280."

Or you can keep it real simple; from jacket's work above:
xt = 280; then 2xt = 560
SO:
560 - 8x = 280
8x = 280
x = 35
 
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