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Homework Help: Find the total distance travelled.

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A car starts at rest and accelerates in 4 seconds to a speed of 20 m/s, which it maintains for the next 8 seconds. Find the total distance travelled.

    2. Relevant equations
    d= vi*t +1/2 at^2
    d= Vavg*t = Vf+Vi/2 *t

    3. The attempt at a solution
    The total time is 12seconds right?
    I can't figure this one out...please help?
  2. jcsd
  3. Sep 26, 2009 #2
    There are two sections, one with an acceleration for 4 seconds reaching 20m/s and the second part is traveling 8s at a constant velocity(20m/s). Find the distance for each section and add.
  4. Sep 26, 2009 #3


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    From rest how far does it travel in the 4 seconds at 20m/s ?

    For the next 8 seconds it travels 20m/s. How far does it travel now?
  5. Sep 26, 2009 #4
    I had two distances, 80 m and 160m, but that sum is not the answer...i'm so confused!
  6. Sep 26, 2009 #5


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    your second distance is correct.

    For the first distance, from rest to 20m/s, it forms a triangle with base of 4 seconds, the area of this triangle gives the distance traveled for this time interval.
  7. Sep 26, 2009 #6
    I found the second distance easy because i had velocity and time. So I just used the formula d= v*t, but i'm curious...our teacher has not shown us the triangle area way of figuring this type of problem out. Our teacher wants us to use a formula with known values where we solve for the unknown. I don't know which formula to use to solve for the unknown first distance. Please advice? Thanks.
  8. Sep 26, 2009 #7
    d= vi*t +1/2 at^2

    Just figure out a. Note: It goes from 0 to 20m/s in 4s.
  9. Sep 26, 2009 #8
    I got the following for a:
    a= (vf - vi/t)
    = (20-0/4)
    =5 therefore a =5m/s^2 Is that the correct value for "a" ?
    then plugging a into d= vi*t +1/2 at^2
    first distance= 0(4) +1/2(5)(4)^2
    = 0 + 1/2 (5)(16)
    = 0 + 1/2 (80)
    = 40
    therefore first distance = 40m Is that the right value for distance?
    The answer in the book says the total distance is 200m, so adding the first and second distances does equal 200, so did i go about this the right way? I guess i did since the answer is the same...hope it's not just a coincidence. Thanks for your help! :-)
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