1. Sep 23, 2012

### supermiedos

1. The problem statement, all variables and given/known data

"A" is visiting "B" on his house. First, "A" starts walking at 2.5 mph, and some time after, he starts running at 5.3 mph. What is the distance "d" from "A" to "B"'s house, if it took 0.5 hrs. to get there?

2. Relevant equations

d = vt

3. The attempt at a solution

I splitted this problem in two parts:

First, I call 'x' the distance he traveled walking at 2.5 mph, and t1 the time it took. So,

x = 2.5*t1 ....(1)

Next, the distance he traveled running at 5.3 mph is d - x, and t2 is the time it took him. So

(d - x) = 5.3*t2...(2)

But, since 0.5 hrs = t1 + t2, we substitute on (2) and get

(d - x) = 5.3*(0.5 - t1) ...(3)

Using eq (1) on (3), I get

d - 2.5*t1 = 2.65 - 5.3*t1
d = -2.8*t1 + 2.65

Which is one equation with two variables. I'm stuck there, can you help me please?

2. Sep 23, 2012

### NasuSama

Then, we know that...

v_1 = 2.5 mph
v_2 = 5.3 mph

Find the average velocity by using this form:

v_avg = (v_1 + v_2)/2

Now, given that t = 0.5 hrs and the formula x = v * t, find the distance.

3. Sep 23, 2012

### supermiedos

Oh, i didn't know you cuold get the average velocity like that. It's because both velocities are constant, right? Thank you so much.

4. Sep 23, 2012

### NasuSama

More like it! Hope this helps! Otherwise, if there are some flaws in my work, let me know!