# Thread: Train Travel... Word Problem

1. ## Train Travel... Word Problem

Hate to do this, but I'm having trouble setting up the following word problem:

A train leaves Union Station for Central Station, 216 km away, at 9 a.m. One hour later, a train leaves Central Station for Union Station. They meet at noon. If the second train had started at 9 a.m. and the first train at 10:30 a.m. they would still have met at noon. Find the speed of each train.

Now I've tried to set it up as

216=(r)1.3 (1)
216=(r)3 (2)

Problem is that; a) I've already set myself up for messing up this problem (because the mph varies too much) b) it gives the trains time up to noon and that makes the 216 km invalid(?).

Thanks

2. Originally Posted by CODEONE
Hate to do this, but I'm having trouble setting up the following word problem:

A train leaves Union Station for Central Station, 216 km away, at 9 a.m. One hour later, a train leaves Central Station for Union Station. They meet at noon. If the second train had started at 9 a.m. and the first train at 10:30 a.m. they would still have met at noon. Find the speed of each train.

Now I've tried to set it up as

216=(r)1.3 (1)
216=(r)3 (2)

Problem is that; a) I've already set myself up for messing up this problem (because the mph varies too much) b) it gives the trains time up to noon and that makes the 216 km invalid(?).

Thanks
for the sake of clarity, let's break this problem up case by case and train by trian

CASE 1: train 1 leaves at 9am and train 2 leaves at 10am

Now they meet at noon, so the time train 1 is traveling for is 3 hours, while the time train 2 is traveling for is 2 hours. we don't know how far they travel before meeting, so say train 1 travels $x$ km, then train 2 will travel $(216 - x)$ km

let the speed of train 1 be $s_1$
let the speed of train 2 be $s_2$

using (again) $\mbox{Speed } = \frac {\mbox{Distance}}{\mbox{Time}}$

we have that for train 1: $s_1 = \frac x3$

and for train 2: $s_2 = \frac {216 - x}2$

CASE 2: train 1 leaves at 10:30am and train 2 leaves at 9am (they are traveling at the same speed as before)

in this case, say train 1 traveled $y$ km and thus train 2 traveled $(216 - y)$ km

so we have (we derived this similar to the way we did in case 1) that the speed of train 1 is: $s_1 = \frac {y}{1.5}$

and the speed of train 2 is: $s_2 = \frac {216 - y}3$

now we equate the formulas for the speeds.

equating the formulas for $s_1$, we get:

$\frac x3 = \frac y{1.5}$ ...................(1)

equating the formulas for $s_2$, we get:

$\frac {216 - x}2 = \frac {216 - y}3$ ..................(2)

those are your simultaneous equations, go get 'em!

3. ## Trouble Again

Alright, well solving the equation gave me the following:

x = 54
y = 108

I've checked them and their right however, the question is asking for "the speed of each train". x and y represent the distances train 1 traveled in Case 1 and Case 2; which case is the question referring to? I'm stuck at this part =/

EDIT: Is it referring to Case 2 since its the case before the "fine the speed of each train". Which would make train 1 travel at 36 km/h and train 2 travel at 54 km/h.

4. Originally Posted by CODEONE
Alright, well solving the equation gave me the following:

x = 54
y = 108

I've checked them and their right however, the question is asking for "the speed of each train". x and y represent the distances train 1 traveled in Case 1 and Case 2; which case is the question referring to? I'm stuck at this part =/

EDIT: Is it referring to Case 2 since its the case before the "fine the speed of each train". Which would make train 1 travel at 36 km/h and train 2 travel at 54 km/h.
Originally Posted by Jhevon
...

let the speed of train 1 be $s_1$
let the speed of train 2 be $s_2$

...

we have that for train 1: $s_1 = \frac x3$

and for train 2: $s_2 = \frac {216 - x}2$

...
..