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Kinematics, Average Speed.

  1. Aug 6, 2015 #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 58 km/h and the other half at 103 km/h. On the way back you travel half the distance at 58 km/h and the other half at 103 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
    V=d/t



    3. The attempt at a solution
    with a help of a friend online but I still don't understand it.
    a.) (58*t/2 + 103*t/2) / t = 58/2+103/2 = (58+103)/2 = 80.5 The problem is I don't know how did "t" gone
    b.) D/(D/2/(58))+(D/2/(103)) = ??
    c.) none, stuck in A and B so can't answer this
    d.) I don't why "zero" is the answer here pls explain

    Help pls
    I kinda suck at Algebra which lead to sucking in Physics too lol
     
    Last edited: Aug 6, 2015
  2. jcsd
  3. Aug 6, 2015 #2
    If it hadn't gone, you couldn't have solved the equation because you have no information about the duration of the trip.

    What is the problem here? This equation is correct, you just have to solve it.

    You have to use the same equation: v=d/t. What is the distance and time if we look at the entire trip? As you don't have exact data about these, you have to express them somehow and hope that the unknown cancels :D.

    What is the difference between speed and velocity?
     
  4. Aug 6, 2015 #3
    Do you know what average speed and velocity is ? What exactly is your doubt ? What do you mean by none stuck ?
     
  5. Aug 6, 2015 #4
    Sorry I meant to say no answer because I got stuck to letter a and b.

    my doubt is I'm lacking knowledge of doing it algebraically.
    especially in letter A I mean how did t has cancel or eliminated
     
  6. Aug 6, 2015 #5
    Letter A. So can you explain how did "t" got cancel or eliminated
    Letter B. Can you lend me a hand of doing it?
    Letter C. I'll deal with later because I need the answer of A and B to answer this
    Letter D. Hmmm none?
     
  7. Aug 6, 2015 #6
    Average speed is defined as the distance travelled in some particular time . Often , when you move , your speed is not constant , and tends to change . So we define average speed , to refer to the average distance travelled per unit time .

    Post edited .
     
    Last edited: Aug 6, 2015
  8. Aug 6, 2015 #7
    Part D - Average velocity is net displacement ( in a certain time interval ) upon time interval considered . What is net diplacement when you leave SA , and then come back to SA ?

    Part C , involves some calculation .

    Hope this helps .
     
  9. Aug 6, 2015 #8
    Okayy I got letter D thankyou!
    Sir problem is I can't do it algebraically especially A look the first post above, I don't know how t got cancel.
     
  10. Aug 6, 2015 #9
    Ok , yes I had made a mistake in my earlier post .
    Corrected - let time taken to reach H be t . First half , with speed v1 and second half with v2 .

    So , for t/2 , car ( I assume ) travels a distance v1*t/2 , and the second t/2 , v2*t/2 .
    So total distance travelled is equal to v1*t/2 + v2*t/2 = ( v1/2 + v2/2 )*t ←taking t out common .

    Now , average speed = ( v1/2 + v2/2 )*t/t ←formula of average speed .
    Can you now cancel t ?

    I hope this is clear enough . If not , feel free to ask what is not .
     
  11. Aug 6, 2015 #10
    Wow thankyou! That was so clear. Thankyou thankyou!! I'll update you again if I have more questions. Good day sir
     
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