Simple problem about distances

  • #1
supermiedos
63
0

Homework Statement



"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?

Homework Equations



d = vt

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?
 

Answers and Replies

  • #2
NasuSama
326
3
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
supermiedos
63
0
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.
 
  • #4
NasuSama
326
3
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.

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

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