1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kinematics problem mixing up distance and time. I'm so confused.

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data
    You drive on Interstate 10 from San Antonio to Houston, half the time at 67 km/h and the other half at 112 km/h. On the way back you travel half the distance at 67 km/h and the other half at 112 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip? (d) What is your average velocity for the entire trip?


    2. Relevant equations

    x = v0t + 1/2at^2

    3. The attempt at a solution

    For a:

    67km/h (.5t) + 112 km/h (.5t) = x

    .5t (67 km/h + 112 km/h) = x

    (x/.5t) = 179 km/h

    (.5) (x/.5t) = 179 km/h (.5)

    (x/t) = 89.5 km/h

    x = 89.5(t)


    I want to say that the time you travel at 67 km/h and the distance you travel that for both directions is the same, but I have a nagging feeling that that isn't correct. What I know is that the distance will be the same so x in x= v0t will be the same. I'd really appreciate some help with this guys, I'm just stumped.
     
  2. jcsd
  3. Aug 30, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    For a, you are almost there once you have written this formula:
    What you should remember is that the "v" in x = v t is the average velocity (it is only the actual velocity if the velocity remains constant over the time t you are looking at).

    In your formula I quoted above, x is the total distance of the trip, and t is the total time. So if you can rewrite this formula to the form
    (something) t = x​
    then the "something" will be your average velocity (compare to v t = x).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook