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Distance covered by a runner

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A runner accelerates from 0 to 15 ft/sec in 2 seconds.
    She goes at a constant v from 2 sec to 10 sec.
    She deccelerates to 0 at the 20 sec mark

    Q: How much distance does she cover?

    2. Relevant equations



    3. The attempt at a solution

    Here is my work:
    From 0 to 2 sec, she accelerates from 0 ft/sec to 15 ft/sec.
    Her average velocity is 7.5 ft/sec, so in 2 sec she covers 7.5 ft. (round to 8)
    From 2 sec to 10 sec, she goes at a constant velocity of 15 ft/sec, so the distance covered is 15ft/sec*8sec, which is 120 feet
    From 10 sec to 20 sec, she decelerates from 15 ft/sec to 0 ft/sec
    Avg velocity of 15ft/sec / 10 sec = 1.5 ft/sec * 10 sec = 15 ft

    8 ft + 120ft + 15ft = 143 ft in 20 sec.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2007 #2
    D =(0.5*2*15)+(8*15)+(0.5*10*15)
    =15+120+75
    =180ft in 20 sec assuming that there is constant acc and deceleration
     
  4. Sep 20, 2007 #3
    OK, I get the first two figures -- but for the decceleration, she goes from 15ft/sec to 0, for a difference of 15 in 10 seconds, so that's an avg velocity of 1.5ft/sec *10 sec is 15 ft.

    Where did I go wrong?
     
  5. Sep 20, 2007 #4
    You are assuming that the velocity is constant(average) which is not the case when accelerating or decelerating.
    I did the question by using a velocity-time graph.
    U could try to do so...
    (i may be wrong...havent done these kind of questions for nearly one year)
     
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