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Thread: displacement, velocity, acceleration

  1. #1
    Gul is offline
    Junior Member
    Oct 2008

    displacement, velocity, acceleration

    An object initially at rest accelerates 10ms^-2 over a distance of 20m, until it reaches a speed of 20m/s. Another object travelling at 25m/s accelerates at 8ms^-2 for the same time period. It is argued that the second object will travel a greater distance under these conditions, and object two will always travel the greater distance when in motion for the same time as object 1. Is this argument valid:

    What I've done:

    Object 1:

    v(0) = 0m/s
    a = 10ms^-2
    v = 20 m/s

    Object 2

    v(0) = 25
    a = 8


    Object one:

    a = dv/dt = 10

    v = 10t + c
    v(0) = 0 , 0 = 10(0) +c, c = 0

    ds/dt = 10t

    s = 5t^2 + c

    Object two:

    a = dv/dt = 8

    v = 8t+c
    v(0) = 25, 25 = 8(0) + c , c =25
    v = 8t + 25

    ds/dt = 8t+25

    s = 4t^2 +c

    I am not sure what to do from here?
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  2. #2
    Member Ruun's Avatar
    Mar 2009
    North of Spain
    Hi Gul:

    Your question should be on the physics subforum, or better in Physics Help Forum!.

    In this problem both are movements with constant acceleration $\displaystyle a_1$ and $\displaystyle a_2$, given.

    The equations you should know in order to solve: $\displaystyle v(t)=v_0+at$ and $\displaystyle v_{f}^{2}-v_{0}^2=2a \Delta x$.

    With the first one you can compute the interval of time that corresponds to both movements, and with the second one the distance traveled by the object 2.
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