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May 23rd, 2000, 06:39 AM
#1
Thread Starter
Lively Member
Okay, i need help with a math problem! FAST.
A car left a garage travelling at 80km/h. 15 minutes later, another car left the same garage, travelling in the same direction at 100km/h. How long will it take for the first car to catch up with the second car?
I got part of the solution done... D1 = the distance the first car travelled and D2 is the distance the 2nd car travelled and r = the rate.
Let t be the time they meet.
D1 = rt
D1 = 80t
D2 = r(t - 15)
D2 = 100(t - 15)
D1 = D2
Therefor, 80t = 100(t - 15)
NOW I'M STUCK!!! SOMEONE PLEASE HELP!!!
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May 23rd, 2000, 09:16 AM
#2
A car left a garage travelling at 80km/h. 15 minutes later, another car left the same garage, travelling in the same direction at 100km/h. How long will it take for the first car to catch up with the second car?
I got it.. I hope...
100 - 20 = 80 ' the 100 km car is going 20 miles more per hour then the 80 km car
80 / 60 = 1 1/3' the car goes 1.33333333etc.. km per second.
15 * 1 1/3 = 20
he went 20 km in 15 minutes..
so hes 20 km ahead.
in another hour the 80 km person would have gone 100km, in an hour. and the 100 km person would have gone 100km in an hour...
therefore, 1 hour is needed for them to both cathup.
good eh? 
[Edited by denniswrenn on 05-23-2000 at 10:18 PM]
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May 30th, 2000, 01:17 AM
#3
Lively Member
Testing Phase
I say we all jump into the car and test it.
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May 30th, 2000, 03:40 PM
#4
PowerPoster
I think we did answer this question in the General sectin rite?
http://forums.vb-world.net/showthrea...threadid=17431
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May 30th, 2000, 06:29 PM
#5
Conquistador
i posted an answer in your other topic, but what year are you in?
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