# Thread: Distance and Time Problem

1. ## Distance and Time Problem

A train leaves a station at noon and travels east at the rate 30 mi/hr. At 2 P.M. of the same day a second train leaves the station and travels south at the rate 25 mi/hr. Express the distance d (mi) between the trains as a function of t (hr), the time the second train has been traveling.

$d = 5\sqrt{61t^2-144t+144}$

I began with:
$d^2 = (30t)^2+(25(t-2))^2$
Am I incorrect?

2. Nevermind... I found a way to the answer!

I believe the correct way to start is:
$d^2 = (30(t+2))^2+(25t)^2$

3. Originally Posted by jesse
A train leaves a station at noon and travels east at the rate 30 mi/hr. At 2 P.M. of the same day a second train leaves the station and travels south at the rate 25 mi/hr. Express the distance d (mi) between the trains as a function of t (hr), the time the second train has been traveling.

$d = 5\sqrt{61t^2-144t+144}$

I began with:
$d^2 = (30t)^2+(25(t-2))^2$
Am I incorrect?
Note quite your model isn't valid for

$0 \le t \le 2$

When t=1 the trains should be 30 miles apart!

4. It's from a book from the 50s. They were still trying to figure this whole math thing out back then.

5. Hi jesse,
At time zero eastbound is 60 miles out .southbound is 0 miles out
after t hrs EB is 60 +30t miles out
SB is 25t miles out
d^2 =(60+30t)^2 + (25t)^2
solve for d in terms of t

bjh