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Homework Help: Help With Problem About 1-D Motion

  1. May 30, 2008 #1
    Okay, so I thought I was on the right track with this problem, but clearly I haven't been getting the right answer!

    The problem is as follows :

    A physics instructor drives down to St. Louis to see his mom. Coming back, he drives the first half the distance at 65 km/hr and the second half of the distance at 100 km/hr. What is his average speed coming back from St. Louis ?

    Okay, so I was trying to determine the t for the two distances and did the following work :

    Well I was going to just post a scan of my work, but it says I can't so I'll just give some of what I did...


    I used t=x/v to solve for the initial t and final t.

    So... t initial = x initial / 65km/hr and t final = x final / 100km/hr

    Then I plugged those into the equation for average velocity = change in x / change in t

    So... V avg = (x - x initial) / [(x / 100) - (x initial / 60)]


    I think I've gotten to the right point, but if the initial distance and final distance are the same I would just get zero, so I know I'm doing something wrong. Any help would be appriciated! :)
  2. jcsd
  3. May 30, 2008 #2
    K, unless I'm very useless at the moment and dumb which happens from time to time... at first, i thought since there's 2 half distances... you add the 2 speeds, and divide them by 2.... giving 82.5km/h but i figured that was wrong.

    The other way I tried was...

    the total distance is 'x' and V_avg = dist/time

    since you dont know time... you have to add in

    V_avg = x / [(dist/2)/V1 + (dist/2)/V2]

    = x / [(x/2)/65 + (x/2)/100 ]

    = 78.78km/h

    For some reason that seems wrong, but I gave it a try.
  4. May 30, 2008 #3
    In your equation, "V avg = (x - x initial) / [(x / 100) - (x initial / 60)]" you also said that the initial distance x and final distance are the same then. If that was so, then the numerator would be zero. In your case, the initial "x_initial" would be 0 and x would be the total distance traveled.

    By definition of average speed, v=(total distance traveled)/(total time traveled)
    Let x = total distance traveled, let x1 = distance of first half, x2=distance of second half, and x1=x2
    Let t = total distance traveled, let t1= time of first half traveled, t2= time of second half
    Let v1=velocity of first half, v2= velocity of second half

    We know that v1=x1/t1, and v2=x2/t2
    Then, 65km/hr=(x1/t1) and 100km/hr=(x2/t2). Since x1=x2, then substitute we get:
    We can solve for t1=(100/65)*t2 (Eq 1.)

    Thus, the average speed would be (x1+x2)/(t1+t2)=(x1+x2)/((100/65)*t2+t2)
    by substituting (eq. 1 for t1). Continue from here, we get
    t2=x/(2*100km/hr) (Eq. 2)
    average speed = x/((100/65)+1)*(x/(200))...
    after some simplification, then average speed = 200/(100/65+1)
    Last edited: May 30, 2008
  5. May 30, 2008 #4


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    lukas86 is showing you the correct way to set it up. Though lukas86 probably shouldn't have given the numerical answer. Your attempt is something like correct also. If xinitial is the distance for the first part of the trip and x is the distance for the last, then xinitial=x. But the total distance isn't x-xinitial. It's x+xinitial. You should be adding the times in the denominator as well.
  6. May 30, 2008 #5
    Ya sorry, I realized that after I posted it. When I looked at it I was like... ya, I'm about the only one that understands what I mean, I posted it fast partially because I'm at work and should be doing... well... work haha. Although physics relates to my job so I thought it would be good to refresh my head a bit. I have written on paper what konthelion has pretty much, but my post was vague and confusing I'm sure. I posted a thread, but either of you familiar with the TI-89 calculator (titanium)?
  7. May 31, 2008 #6
    Thanks you guys! I kept making xinitial 1/2 of xfinal and was getting the wrong answer...besides the fact that I was calculating the change in distance/change in time when I now realize that it's really just the total trip thanks to you lukas! I knew the x value for both velocities had to be the same, but I just couldn't grasp it at first haha

    Thanks again and I'm sure I'll be back soon enough with more questions! :)
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