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Finding distance given two different velocities and times?

  1. Oct 16, 2014 #1
    1. The problem statement, all variables and given/known data
    On a straight track, a cart travels with a constant velocity of 2.5m/s for 9.0s. It then uniformly accelerates to a final velocity of 6.0m/s in 15s. Calculate the total distance travelled by the cart.

    2. Relevant equations
    Vavg = Vi+Vf/2
    d=(Vf+Vi/2)x t
    d=Vi x t + 1/2at^2
    a=Vf-Vi/t
    3. The attempt at a solution
    I tried finding the average velocity by adding together 6.0 and 2.5 before dividing it by 2 to get 4.25. Then, I multiplied that number by the total time of 24s (15+9). This got me the number 102, but the answer key says the answer is 86?
     
  2. jcsd
  3. Oct 16, 2014 #2

    SteamKing

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    IDK why you chose this approach. The problem clearly states that the cart accelerates from 2.5 m/s to a final velocity of 6.0 m/s, which means that during the acceleration, average velocities will not give correct distances. You state the proper formulas in your post; why not use them?

    You need to work this problem in two phases: 1. the portion where the velocity is constant, and 2. the portion where the car is accelerating.
     
  4. Oct 16, 2014 #3

    DaveC426913

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    More succinctly, why did you multiply the average speed of the second (accelerative) leg of the journey by the duration of the entire journey?
     
  5. Oct 16, 2014 #4
    The formulas you are using are for constant acceleration, but there isn't constant acceleration overall. There is zero acceleration for the first 9.0 s, and then there is non-zero acceleration from then until the end of the problem. You cannot apply these formulas over the entire interval in question, but you can apply them to each interval alone. This is one way of doing the problem.

    Perhaps a more intuitive and easier-to-see approach would be to graph the velocity as a function of time (you'd have a piece-wise graph) and then integrate your graph over the entire interval (you'd have to do two integrals). This would amount to finding the area under the graph, and the total area under the graph will be the total displacement.
     
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