Distance, Rate, and Time - Word Problem
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Distance, Rate, and Time - Word Problem

[From: ] [author: ] [Date: 12-01-17] [Hit: ]
5) - 20t(t+2.t = -7.5 or 5 hours : discard the negative value since the train isnt traveling in reverse.The freight train makes the trip in 7.......
I know I have to use Distance rate and time, but I keep getting a super large number. Here is the problem:
A freight train requires 2 1/2 hours longer to make a 300 mile journey than an express train does. If the express averages 20 MPH faster than the freight train, how long does it take the express train to make the trip?

So, i have distance= 300, how do i figure out the rate? I got rate of freight = x - 20 ; and the rate of express= x
Is that correct? and then that means the time of freight= 300/x-20 ; and the time of express= 300/x
What do I do next with the 2 1/2 hours?
is it "time of express is 2 1/2 hours less than the time of the freight" which means:
300/x = (300/x-20) - 2 1/2 ?

If I am doing it wrong PLEASE tell me, I really don't want to be doing it wrong especially since I have a quiz tomorrow! Any input would be wonderful.
Thanks in advance.

And yes, I posted this earlier, but I never got a response!!

-
t = time for the express train to make the trip
speed = distance/ time
distance = 300
time of the freight = t+2.5
speed of the freight, 300/(t+2.5) is speed of the express minus 20, 300/t - 20

300/(t+2.5) = 300/t - 20
300t = 300(t+2.5) - 20t(t+2.5)
300t = 300t + 750 - 20t² - 50t

20t² + 50t - 750 = 0
2t² + 5t - 75 = 0
(2t + 15)(t - 5) = 0
t = -7.5 or 5 hours : discard the negative value since the train isn't traveling in reverse.
t = 5 hours

The express train makes the trip in 5 hours at 60 mph
The freight train makes the trip in 7.5 hours at 40 mph
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