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Physics average speed

  1. Jan 24, 2005 #1
    Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 50.0 and half the distance at 76.0. On her return trip, she drives half the time at 50.0 and half the time at 76.0.

    they are looking for average speed to grannys house and then from grannys house.

    can someone give me a push in the right direction, im sufficiently lost.
  2. jcsd
  3. Jan 24, 2005 #2


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    Average speed is given by the total distance travelled (100 miles), divided by the total time. So, calculate the time taken to travel the first 50 miles at the speed given, and the time taken to travel the second 50 miles at the speed given, and add those two times together to get the total time for the trip to Granny's house. 100 divided by this time is the average velocity there.

    Do this again for the way back.
  4. Jan 24, 2005 #3
    well for the way back i got 63 as the correct answer through guessing. 63 was nearly correct for the way there.

    well in my head i can see that the first half will be 30 minutes. by calculate im lost. im still not understanding much of the equations in this physics book, or if there are physics equations?
  5. Jan 24, 2005 #4
    just got answer of 60, dont ask how.
  6. Jan 24, 2005 #5


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    It was a lucky guess. You assumed that the average speed would be the average of the two speeds given, which is true only because she drove for the same amount of time at each speed. The two velocities therefore had the same *weighting* in the average.

    The average of the two velocities: (50 + 76) / 2 = 126/2 = 63 mph

    Average velocity according to the definition distance / time:

    = [50(t1) + 76(t2)] / t_total

    The problem tells us that t1 = t2 = t (she drives at each spped for the same amount of time). So obviously t_total = 2t So the equation becomes:

    [50(t1) + 76(t2)] / 2t = 126t / 2t = 63

    In this case, the two methods give the same result. But what if she had not driven at each speed for the same amount of time? For instance, if she drove at 76 mph for LONGER than she drove for 50 mph, obviously the average speed would be higher than 63, and would be closer to 76 than to 50. 76 would have a higher *weighting* in the average. If she had driven for longer at 50mph than at 76mph, then her average speed would have been lower, closer to 50 than to 76. This is the case on the way TO Granny's house. She does not drive at each speed for the same amount of time. t1 is not equal to t2. Use the method I told you about in my first post to find each of these times, and solve for the average speed.

    Huh? Are you sure? Are you talking about on the way there? Because if she drives halfway (50 miles) at 50 miles PER HOUR, then it's obviously going to take her a full hour to do so.
    Last edited: Jan 24, 2005
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