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Basic Kinematics question

  1. Feb 2, 2007 #1
    A car traveled up a hill at constant speed of 10.0 m/s and then returned down the hill at 20.0 m/s . If the time to turn around is ignored, what was the average speed for the trip?

    Now i like to struggle to figure out questions myself but i cant seem to figure this one out. I have most of the kinematics formulas in front of me too. If anyone could help to give me a hint or a clue it would be great. thanks an advance.
  2. jcsd
  3. Feb 2, 2007 #2

    Doc Al

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    Staff: Mentor

    Try something! What's the definition of average speed? Hint: Call the distance traveled "D".
  4. Feb 3, 2007 #3
    so while going down the hill would the distance then become 2D?
  5. Feb 3, 2007 #4

    Doc Al

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    If the one-way distance is D, the round trip is 2D.
  6. Feb 3, 2007 #5
    ok i got the answer but i do not think that i did it the right way. please take a look at my steps

    Average Speed = [Distance Traveled]/[Time Taken to Travel that Distance]

    1. I pretend the distance travelled one way up the hill is 10m, therefore 20m round trip.

    2. I then figure out the average time it takes to go up and down the hill at 10m/s and and at hill at 20m/s. (20/10+20/20)/2=1.5s

    3. I then use the average time and roundtrip distance to find average speed. 20/1.5=13.33m/s

    Is this the correct way of solving the problem? Is my step 2 the correct way of thinking?
  7. Feb 3, 2007 #6
    You were correct in saying that the distance would be defined as 2D.

    So now we must define time. We know that 2D must be divided by time, but how do we define time? Simple, were change the formula to define it as

    2D all over D/v going up + D/v going down.

    Do some cancellations and then plug and chug.
  8. Feb 4, 2007 #7

    Doc Al

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    That's OK. You could also stick with calling the distance D--you'll find it doesn't matter, the distance will drop out.
    Think of it this way:
    How much time to go up? t1 = 10/10 = 1 s
    How much time to go down? t2 = 10/20 = 0.5 s
    Total time = t1 + t2 = 1.5 s

    Right. Total distance (20) over total time (1.5) equals average speed.

    See my comments above.

    Since the actual distance doesn't matter, picking a specific distance (like 10 m) is OK. But doing it algebraically is cooler:
    How much time to go up? t1 = D/10
    How much time to go down? t2 = D/20
    Total time = t1 + t2 = D/10 + D/20 = D(1/10 + 1/20) = D(3/20)

    Total distance = 2D
    Average speed = 2D/[D(3/20)] = 2/(3/20) = 40/3 = 13.33 m/s
    (Note how the Ds cancel!)
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