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Average Speed and Velocity

  1. Nov 4, 2008 #1
    Im having a rough time with this set of problems in general, so instead of creating a lot of threads im just going to keep adding what I have here. I am not looking for the answers to my homework, just some guidance and my mistakes pointed out. I appreciate any and all input I receive.

    1. The problem statement, all variables and given/known data

    You jog at 9.5km/h for 8.0km, then you jump into a car and drive an additional 16km. With what average speed must you drive your car if your average speed for the entire 24km is to be 22km/h?



    2. Relevant equations

    I typed the entire question, so im not sure what this specifically wants.



    3. The attempt at a solution

    I calculated the time jogging to be 8km/9.5km/h = .8421 hrs.
    Then the amount of time to travel 24km with an average speed of 22km/h (22km/h)/(24km) = 0.9167
    Next I subtracted the time jogging from the total time needed to travel, to determine how long the car had to cover the remaining distance. 0.9167 - 0.8421 = 0.0746.
    and found the average speed of the car using average speed = distance / elapsed time
    16km/0.0746hrs. = 214.4772km/h.
    I think with significant digits I report 214.5km/h as my answer for the speed the car must travel.

    That answer doesnt seem very realistic to me, so I am doubting my process. Also I think theres a much more elegant way or formula to use to solve this problem. Any input?
     
  2. jcsd
  3. Nov 4, 2008 #2
    The average speed for the entire trip is the weighted average of the speeds of jogging and driving. Set that up as an equation:

    (9.5*8+x*16)/24=22
    x = 28.25 km/h
     
  4. Nov 5, 2008 #3
    This would be true only if by "average speed" one means "distance-averaged" speed, which I'm sure is not the purpose of the question.
     
  5. Nov 5, 2008 #4
    You went wrong in this last step: clearly, if you travel 24 km with an average speed of 22km/h, you're gonna take more than 1 hour, not less!
     
  6. Nov 5, 2008 #5
    but isnt average speed found by time/distance? I see the logic in taking more than one hour, not less. But I dont understand exactly how I messed that up. I used the formula for average speed from my book and that was the output.
     
  7. Nov 5, 2008 #6
    Read my second post first, but for other people's purpose I'll leave this.

    Just stopping by these forums for some reason, but it looks like you need some help.

    I will go through it conceptually for you, and then give you some equations.

    First we have to remember what we are looking for. We are looking for the avg. speed the car must travel, so we know that in the end- we will need the distance it travels, and the time it took.

    I'd do this problem in steps. The first one is to see how long the actual trip was. We do this so we can subtract the amount of time it took to walk from the total to see how long the car trip was.

    Given 24km trip, and 22 km/h, it is a trivial problem (Vavg=D/T) to find the time. I'll call this Ttotal (Tt).

    Now we have the time it took to walk, you have correctly found this value to be .8421 hours. Subtracting this value from Tt will give you the time in which the car was travelling.

    You now have every component you need to solve for the average velocity of the car.

    V(car avg) = Distance by car / time by car

    V= 16km / (Tt-.8421)

    I could do this problem in one step, but I find it's easier for most people to do these problems sequentially and their answer has a higher fidelity.
     
    Last edited: Nov 5, 2008
  8. Nov 5, 2008 #7
    Just looked at your first post, and you had the method right. You made a trivial mistake in calculating the total time.

    It's Vavg= D/T or 22= 24 / T which means T= 24/22 or 1.09 something.

    Might as well ignore my above post because you got it right besides that :)
     
  9. Nov 5, 2008 #8
    You are kidding right? Speed is time/distance????
     
  10. Nov 5, 2008 #9
    as = distance / elapsed time.
    sorry was looking at too many things at once.
     
  11. Nov 5, 2008 #10
    I think I resolved this as 62km/h. Thanks chuck.
    I used t=24/22 = 1.1 - .8421 = .2579
    16/.2579 = 62km/h.

    thanks all for help.
     
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