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A box sliding down a incline plane

  1. Sep 20, 2009 #1
    A 3kg block starts from rest at the top of a 30degree incline and slides a distance of 2m down the incline in 1.5s find the magnitude of the acceleration.

    I have always though to find the acceleration you can just use x/(t*t) but i use this formula but i couldn't get the same as the book. is there something wrong with my understanding?
     
  2. jcsd
  3. Sep 20, 2009 #2
    Your formula is incorrect.

    You need to use the basic kinematic equation for motion under constant acceleration here. If you happen to need anything else, just differentiate it with respect to time
    [tex]x(t)=x_0+v_0t+\tfrac{1}{2}at^2[/tex]

    Consider the case where the mass starts at rest, with [tex]x_0=0[/tex]
     
  4. Sep 20, 2009 #3

    rl.bhat

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    Homework Helper

    Write down the relevant equations.
    Show your calculations.
     
  5. Sep 20, 2009 #4
    can you explain why cant i just use x/(t*t) ?
     
  6. Sep 20, 2009 #5
    x/(t*t) is the units of acceleration. For example meters per second squared

    [tex] \frac {m/s^2} [/tex]
     
  7. Sep 20, 2009 #6
    It is not, however, the formula for the distance traveled. It may share the same units, but it's off by a factor of two.

    Let's begin with the basics. We want to develop the kinematic equation for one dimensional motion under constant acceleration:

    Our only givens are the initial position with respect to some origin, the initial velocities and accelerations with respect to a certain frame of reference, and the fact that the acceleration is constant.

    [tex]a(t)=constant[/tex]
    Taking the indefinite integral over time:

    [tex]v(t)=a\cdot t + C[/tex]

    [tex]C=v_0[/tex]

    [tex]v(t)=a\cdot t+v_0[/tex]

    We have now found an expression for the velocity as a function of time.

    Taking the indefinite integral over time once more yields:

    [tex]x(t)=\tfrac{1}{2}a\cdot t^2 +v_0\cdot t + C[/tex]

    [tex]C=x_0[/tex]

    [tex]x(t)=\tfrac{1}{2}a\cdot t^2 +v_0\cdot t + x_0[/tex]

    Q.E.D.

    Just to elaborate on the point that has you confused though, differentiate the expression [tex]\tfrac{1}{2}at^2[/tex] with respect to time.
    Remember, [tex](ax^n)'=anx^{n-1}[/tex]
     
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