Equations of motion (falling body)
- Thread starter titeri47
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topsquark
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The formula you are using implies that the initial position is at the origin and the object was at rest at this time. (So [math]d_0 = 0[/math] and [math]v_0 = 0[/math]. Further I'm going to assume that t = 0 at this point.)
The easiest way is to use Calculus.
[math]a = \dfrac{dv}{dt} \implies v = \int_0^t a ~ dt[/math]. If a is constant then we have [math]v = at[/math].
[math]v = \dfrac{dx}{dt} \implies x = \int_0^t v ~ dt = \int_0^t at ~ dt = \dfrac{1}{2} at^2[/math]. (I'm using x for distance due to the possible confusion of the dt's floating around.)
Non-Calculus isn't too bad, but you have to know a somewhat unused equation.
For a constant acceleration we know that
[math]x = \frac{1}{2}vt[/math]. (This is one of the four constant acceleration equations [math]x = \dfrac{1}{2}(v_0 + v)t[/math], where [math]v_ 0 = 0[/math] in this case.)
Then v = at gives [math]x = \dfrac{1}{2} (at) t = \dfrac{1}{2}at^2[/math]
-Dan
The easiest way is to use Calculus.
[math]a = \dfrac{dv}{dt} \implies v = \int_0^t a ~ dt[/math]. If a is constant then we have [math]v = at[/math].
[math]v = \dfrac{dx}{dt} \implies x = \int_0^t v ~ dt = \int_0^t at ~ dt = \dfrac{1}{2} at^2[/math]. (I'm using x for distance due to the possible confusion of the dt's floating around.)
Non-Calculus isn't too bad, but you have to know a somewhat unused equation.
For a constant acceleration we know that
[math]x = \frac{1}{2}vt[/math]. (This is one of the four constant acceleration equations [math]x = \dfrac{1}{2}(v_0 + v)t[/math], where [math]v_ 0 = 0[/math] in this case.)
Then v = at gives [math]x = \dfrac{1}{2} (at) t = \dfrac{1}{2}at^2[/math]
-Dan