1. rate*time=distance

A passenger train and a freight train start at 405 miles apart and travel towards each other. the rate of the passenger train is twice the rate of the freight train. in 3 hours they pass each other (or collide). how fast are they going?

A passenger train and a freight train start at 405 miles apart and travel towards each other. the rate of the passenger train is twice the rate of the freight train. in 3 hours they pass each other (or collide). how fast are they going?
Pet Peeve Warning!!! Danger Will Robinson! Danger! Danger!

Why can't Math books just use the term "speed" like everyone else? "rate*time = distance." "rate" of what, I ask you???

Pet Peeve over.

Okay. Let's assume that the passenger train starts from the left and the freight train starts from the right. For the record I am going to set an origin at the starting position of the passenger train and set the +x direction to be to the right.

The passenger train will move with a speed $\displaystyle v_p$ and cover a distance x miles. The freight train will move with a speed $\displaystyle v_f$ and covers a distance 405 - x miles. They meet at the same time, t = 3 hours.

So for the passenger train:
$\displaystyle x = v_pt$

and for the freight train:
$\displaystyle 405 - x = v_ft$

and
$\displaystyle v_p = 2v_f$

Put the first equation into the second:
$\displaystyle 405 - (v_pt) = v_ft$

and now insert the 3rd equation into the equation we just got:
$\displaystyle 405 - (2v_f)t = v_ft$

Now we can solve for $\displaystyle v_f$:
$\displaystyle 405 - 2v_ft = v_ft$

$\displaystyle 405 = 3v_ft$

$\displaystyle v_f = \frac{405}{3t} = \frac{405 \, mi}{3 \cdot 3 \, h} = 45 \, mi/h$

So $\displaystyle v_p = 2v_f = 90 \, mi/h$

-Dan

A passenger train and a freight train start at 405 miles apart and travel towards each other. the rate of the passenger train is twice the rate of the freight train. in 3 hours they pass each other (or collide). how fast are they going?
Let s be the speed of the frieght train.
Then the speed of the passenger train is 2s.

The average speed of both trains is:
(s+2s)/2=(405/3)/2 <---- 405 / 3 = distance / time
(3s)+2=135/2
1.5s = 67.5

The speed of the frieght train:
1.5s = 67.5
s = 67.5 / 1.5
s = 45

The speed of the passenger train:
1.5s = 67.5 <---- solving for 2s
s = 45
2s = 90

The speed of the freight train is 45 mph.
The speed of the passenger train is 90 mph.

4. thx

thx! i get it now!