One Dimensional Motion Derivations
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Derivation of the four 1-D motion equations under constant acceleration.
Contents |
[edit] The Average Velocity
For 1-D motion under constant acceleration we have the equation (1) <math>\bar{v} = \frac{1}{2}(v_0 + v)<\math>
Please do not make the mistake of assuming that the average velocity is generally the arithmatic mean of the initial and final velocities!! This is not a general equation.
[edit] The Velocity Equation
<math>a = \frac{\Delta v}{\Delta t}<\math>
<math>a = \frac{v - v_0}{t} (Recall that we are assuming <math>t_0 = 0<\math>.)
<math>at = v - v_0<\math>
<math>v = v_0 + at<\math>
[edit] The (Less Common) x Equation
The next simplest equation to derive is the one that is probably used the least. We have the definition of the average velocity <math>\bar{v} = \frac{\Delta x}{\Delta t}<\math>
From equation (1) above: <math>\frac{1}{2}(v_0 + v) = \frac{x - x_0}{t}<\math>
<math>\frac{1}{2}(v_0 + v)t = x - x_0<\math>
<math>x = x_0 + \frac{1}{2}(v_0 + v)t<\math>
[edit] The x Equation (Calculus Derivation)
We start with the definition of the instantaneous acceleration <math>a = \frac{dv}{dt}<\math>
Integrating the acceleration equation with respect to time gives us <math>dv = a~dt<\math>
<math>\int_{v_0}^v dv = \int_0^t a~dt<\math>
Since we are considering the acceleration to be a constant we may move it outside the integral. <math>v - v_0 = a \int_0^t dt = a(t - 0) = at<\math>
So <math>v = v_0 + at<\math> as we already knew.
Now we integrate this equation with respect to time. To do this, note that <math>v = \frac{dx}{dt}<\math>
<math>\frac{dx}{dt} = v_0 + at<\math>
<math>dx = (v_0 + at)dt<\math>
<math>\int_{x_0}^x dx = \int_0^t(v_0 + at)dt = v_0t + \frac{1}{2}at^2<\math>
So <math>x = x_0 + v_0t + \frac{1}{2}at^2<\math>
[edit] The x Equation (Non-Calculus Derivation)
We start with the previously derived (less common) x equation <math>x = x_0 + \frac{1}{2}(v_0 + v)t<\math>
and recall the velocity equation <math>v = v_0 + at<\math>
Inserting this into the x equation gives <math>x = x_0 + \frac{1}{2}(v0 + (v_0 + at))t<\math>
<math>x = x_0 + \frac{1}{2}(2v0 + at)t<\math>
<math>x = x_0 + v_0t + \frac{1}{2}at^2<\math>
[edit] The v Squared Equation
Start with the (less common) x equation <math>x = x_0 + \frac{1}{2}(v_0 + v)t<\math>
<math>x - x_0 = \frac{1}{2}(v_0 + v)t<\math>
(2) <math>2(x - x_0) = (v_0 + v)t<\math>
Now, take the velocity equation and solve it for t <math>v = v_0 + at<\math>
<math>at = v - v_0<\math>
<math>t = \frac{v - v_0}{a}<\math>
Insert this value of t into equation (2) above. <math>2(x - x_0) = (v_0 + v) \frac{v - v_0}{a}<\math>
<math>2(x - x_0) = \frac{(v + v_0)(v - v_0)}{a}<\math>
<math>2a(x - x_0) = (v + v_0)(v - v_0)<\math>
<math>2a(x - x_0)= v^2 - v_0^2<\math>
<math>v^2 = v_0^2 + 2a(x - x_0)</math>
