1. ## distance problem

I have having problem solving this.
One train leaves a station heading due west. Two hours later a second train leaves the same station heading due east. The second train is traveling 15 mi/hr faster than the first. Six hours after the second train leaves, the two trains are 580 miles apart. Find the rate at which east train is traveling.

2. Originally Posted by barefoot12
I have having problem solving this.
One train leaves a station heading due west. Two hours later a second train leaves the same station heading due east. The second train is traveling 15 mi/hr faster than the first. Six hours after the second train leaves, the two trains are 580 miles apart. Find the rate at which east train is traveling.

$\displaystyle 1st=x$

$\displaystyle 2nd= 15+x$

$\displaystyle total in 6 hrs= 6(x+15+x)=580$

$\displaystyle 12x+90=580$

$\displaystyle 12x=490$

$\displaystyle x=40.83 mi/hr$

$\displaystyle 15+x= 15+40.83$

$\displaystyle 2nd train =55.83mi/hr going east$

3. Originally Posted by barefoot12
One train leaves a station heading due west. Two hours later a second train leaves the same station heading due east. The second train is traveling 15 mi/hr faster than the first. Six hours after the second train leaves, the two trains are 580 miles apart. Find the rate at which east train is traveling.
Try setting up the information explicitly in terms of the "d = rt" equation.

. . . . .first train:
. . . . .rate: r
. . . . .time: 2 + 6 = 8
. . . . .distance: 8r

. . . . .second train:
. . . . .rate: r + 15
. . . . .time: 6
. . . . .distance: 6(r + 15) = 6r + 90

They are headed in opposite directions, so the distance between them is the sum of their two distances:

. . . . .total distance:
. . . . .8r + 6r + 90 = 580

Then:

. . . . .14r = 490

Divide through by 14 to find the rate of the first train. Back-solve to find the rate of the second train.

Have fun!