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?
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 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.

 Quote by 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.
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.