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Problem with Velocity

  1. May 26, 2005 #1
    The initial speed of a car is 25ms^-1 and it accelerates for 4 secs to reach a speed of Vms^-1. It then maintains this velocity travelling at constant speed for a further 8 seconds. The total overla distance travlled is 600m.

    I am really stuck.
    I used v = u+at to get one question.

    v = 25 + 4a......1
  2. jcsd
  3. May 26, 2005 #2
    Ok there are basically two ways to this problem . One is that you have to apply kinematic equations for three different cases.

    Secondly, a better one is that you draw a v-t graph which will start from 25 m/s and with +ve slope , then it goes forward at the same velocity and 0 slope for 8 secs.Now area under this graph is the distance travelled , that is 600 m.

    I think you havent mentioned what is to be found in this problem, but the graph method will give you any value you want.
  4. May 26, 2005 #3
    oh yeah. I have to find the value V. Using the graph method, V comes out to be 75. Right?
  5. May 26, 2005 #4
    If you have correctly drawn the graph , using simple science you will get the correct answer.Check your answers .
  6. May 26, 2005 #5
    I used the trapezium rule for area under the graph (a+b/2 * height) to find V.
    v= 75ms^-1
  7. May 26, 2005 #6
    Yes answer is right then.Good Job.
  8. May 26, 2005 #7


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    Homework Helper

    That is not the correct answer. At 75m/sec, in 8 sec the car would move 600m. The problem says the car moves a total of 600m during the entire time of 4 seconds while accelerating plus 8 additional seconds at constant speed.
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