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Average Speed Question

  1. Oct 10, 2006 #1
    On your wedding day you leave for the church 36 minutes before the ceremony is to begin, which should be plenty of time since the church is only 9.0 miles away. On the way, however, you have to make an unanticipated stop for construction work on the road. As a result, your average speed for the first 18 minutes is only 5.0 mi/h. What average speed do you need for the rest of the trip to get you to the church on time?

    I have tried to complete this problem several times, and I have failed, it is getting pretty frustrating, here is what I have done.

    Average Speed = [Distance][Time]

    18 minutes / 5 mi/h = 3.6 miles traveled?

    36-18 = 18 minutes more to travel 5.4 miles!





    Now there is 5.4 miles left to be traveled. How in the heck do I figure out what my average speed needs to be to travel this within 18 minutes!?
    Im seriously confused and lost, please help!
     
    Last edited: Oct 10, 2006
  2. jcsd
  3. Oct 10, 2006 #2
    The correct formula is average velocity = distance/time.
    Also remember to keep everything in the same units. Either make everything hours or everything minutes.
     
  4. Oct 10, 2006 #3
    Still Lost!

    Im still lost, even if Average Velocity = [Distance][Time] how do I figure out how fast I have to travel within the given time? if what you said is true, I can convert 18 minutes to hours by dividing 18/60 which gives me .3

    .3hrs x 5.0 m/h = 1.5mi traveled?

    So now I have 7.5 miles left to travel within 18 minutes, or

    7.5 miles left to travel within .3 hours, how do I figure out the average speed required for me to get to the church ontime?
     
  5. Oct 10, 2006 #4
    Well... if average velocity = distance/time. We know that he have to travel 7.5 miles, we know that we must travel that in 18 mins (.3 hours) then we can just plug it into the equation

    V = (7.5mi)/(.3hours)
    V = 25 mi/h
     
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