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Velocity and acceleration of a runner

  1. Mar 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Dan starts from rest and started running for a distance of 6.0 x 101 m in 9 s.
    a) What was Dan’s final velocity at this time?
    b) What was the acceleration of Dan?


    2. Relevant equations

    a= final velocity - initial velocity / change in time


    3. The attempt at a solution

    Initial velocity = 0
    t= 9 s
    distance = 60 m


    a= final velocity - initial velocity / change in time

    the problem Im having is that i don't have acceleration or final velocity, so i don't know how i'm sopposed to plug into the formula

    I was just hoping for a push in the right direction! thanks
     
  2. jcsd
  3. Mar 9, 2007 #2

    radou

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    Dan is accelerating uniformly. Which relation tells us about the distance traveled in this case of motion?
     
  4. Mar 9, 2007 #3

    Dick

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    You should probably assume he is accelerating at a constant rate during this time. Do you know any equations relating distance to acceleration?
     
  5. Mar 10, 2007 #4
    distance = (initial velocity)(change in time) + 1/2a (change in time)(change in time)

    that is an equation relating to distance nd acceleration
     
  6. Mar 10, 2007 #5

    Dick

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    Ok, so zero initial velocity. You have the distance, you have the change in time. So you should be able to find 'a', right?
     
  7. Mar 10, 2007 #6
    yes, i think i actually figured it out!

    his acceleration would be 1 because v/t is 15/15 which is 1 m/s squared
    and then you plug that in and his final velocity would be 22.9 but with sigdigs it would be 23

    does that seem right?
     
  8. Mar 10, 2007 #7

    Dick

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    Whoa! Slow down. I now have:

    60 m=(1/2)*a*(9 sec)^2 and I'm asking myself what is 'a'? I don't see any v/t and I don't get 1.
     
    Last edited: Mar 10, 2007
  9. Mar 10, 2007 #8
    Okay, i got excited and started using my givens from a different questions, what i meant was:

    you have to isolate for a so you have

    distance/aceleration = (1/2)(81)
    = 40.5
    then you cross multiply

    distance = acceleration (40.5)
    distance/ 40.5 = a
    60/40.5 = a
    a = 1.481 m/s squared
     
  10. Mar 10, 2007 #9

    Dick

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    Well done. So final velocity is?
     
  11. Mar 10, 2007 #10
    v = at
    = (1.481 m/s)(9)
    = 13.32 m/s
    = sig digs 13 m/s
     
  12. Mar 10, 2007 #11

    Dick

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    Seems fine to me. But I would write (1.481 m/s^2)*(9 s)=13.32 m/s. Keep the units straight.
     
  13. Mar 10, 2007 #12
    Thankyou so much

    Do you think i need to have a direction since it's velocity? or if i say +13 m/s will that be fine?

    Would you help me with one more question? it is

    Toronto downtown and Scarborough Town Centre are 20.0 km apart. A cyclist leaves Toronto and heads for Scarborough at 20.0 km/h. A second cyclist leaves Scarborough for Toronto at exactly the same time at a speed of 15.0 km/h.
    a) Where will the two cyclist meet between Toronto and Scarborough?
    b) How much time passes before they meet (in minutes)?

    I know the distance and two speeds and i am looking for T I was thinking maybe to useing a formula with distance velocity and time, and then isolating for time, but i don't know which formula
     
  14. Mar 10, 2007 #13

    Dick

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    13 m/sec is fine. We are pretty clear on the direction, right? Ok, for the second one just use d=v*t. Let t be the unknown meeting time. What distance does each cyclist travel in time t? When they meet, what will the sum of those distances be? Can you solve the resulting equation for t?
     
  15. Mar 10, 2007 #14
    Okay so you have t = d/t
    t = 20 km/ 15 km/h
    t = 1.33 hours

    So they will meet after 1.33 hours
     
  16. Mar 10, 2007 #15

    Dick

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    Nope. Slow down again. If the time is T, how far does the first cyclist travel?
     
  17. Mar 10, 2007 #16
    I don't know the question doesn't say.d = vt
    = 20 km/h time

    I don't know what to doooo
     
  18. Mar 10, 2007 #17
    I don't know the distance,they don't tell you, and there is no way to find out. It there?
     
  19. Mar 10, 2007 #18

    Dick

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    As you said, the first bicyclist travels (20km/h)*T. The second one travels (15km/h)*T. Since T is the meeting time, if I add those two together, what do I get?
     
  20. Mar 10, 2007 #19
    displacement? so the read formula would be

    t = distance 2 -distance 1 /velocity 2 -velocity 1

    = 20km / 20km/h - 15 km/h
    = 20km/5km/h
    = 4 hours??
     
  21. Mar 10, 2007 #20

    Dick

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    Forget formulas! Answer my question! I want a number of km! What is the sum? How much distance did the riders cover together?
     
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