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Kinematics - particle acceleration and distance travelled

  1. Mar 5, 2014 #1
    Kinematics -- particle acceleration and distance travelled

    1. The problem statement, all variables and given/known data
    A particle moving at 10 m/s reverses its direction to move at 20 m/s in the opposite direction. If its acceleration is -10 m/s2, what is the total distance that it travels?

    A) 15 m

    B) 20 m

    C) 25 m

    D) 30 m

    2. Relevant equations
    a = Δv/t

    v = Δx/t




    3. The attempt at a solution

    In one direction:

    time = velocity/acceleration = 10/10 = 1 s

    velocity = distance/time = velocity * time = distance

    distance = 10 m/s * 1s = 10 m

    In the reverse direction:

    acceleration = velocity/accelertion; time = velocity/acceleration

    time = 20 m/s/10 m/s2 = 2 s

    velocity = distance/time; distance = velocity * time = 20 m/s * 2 = 40 m

    Distance traveled:

    40 m + 10 m =50 m

    I realize my final answer did not come out to be any of the choices above but am not sure of any other correct way to approach this problem.
     
  2. jcsd
  3. Mar 5, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello.

    Your results for the two times look good.

    The equation Δx/t (or Δx/Δt) is the formula for average velocity vavg.

    So, Δx = vavg Δt.

    What are the average velocities for the first and second parts?
     
  4. Mar 6, 2014 #3
    I would use the Kinematic equations, in particular x = V_o(t) + 1/2 at^2 and V_f = V_o + at. The first equation gives you the distance traveled, once you know the time, which is calculable from the other equation. Using this, the answer comes out as one of the choices you listed. This form of the equations is a little easier to work with in my opinion.
     
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