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College physics problem.

  1. Feb 27, 2012 #1
    1.
    Starting from the origin, a car accelerates on a level road from rest at a(A)=-2.00 m/s^2 until T(1)=20.0s. The velocity is then held constant until T(2)=40.0s, and then the car accelerates at a(B)=5.00m/s^2 until T(3)=50.0s.

    question: What is the final position of the car?




    2. delta X



    3. got -400m. but the answer should be -1350m.
     
  2. jcsd
  3. Feb 27, 2012 #2
    Draw a v-t graph, area under the graph is the displacement.
    But indeed I got -1337.5 m
     
  4. Feb 27, 2012 #3

    PeterO

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    Homework Helper

    20 seconds of acceleration achieves -40 m/s; so avergae vel -20.
    -20 x 20 = -400 m while accelerating

    A further 20 seconds at -40m/s covers a further -800m - that's -1200 so far.

    We now accelerate at +5, so it takes 8 seconds to stop.
    Average velosity -20, time 8 sec means a further -160m ; thats -1360m so far.

    The acceleration continues for a further 2 seconds - reaching a velocity of +10 m/s.
    Thats an averag of +5 m/s for 2 seconds, so a displacement of + 10m

    -1360 + 10 = -1350

    SO how did you get your -400 m ? [or -1337.5 for that matter]
     
  5. Feb 27, 2012 #4
    Oh yes, should be -1350 m, cal a wrong time interval for deceleration....
     
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