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The time rate of change of velocity of an object, or alternately the net force on the object divided by the object's mass. |
Mathematical Representation
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\begin{large}\[ \vec{a} = \frac{d\vec{v}}{dt} \qquad \mbox{or} \qquad \vec{a}=\frac{\sum \vec{F}}{m} \]\end{large} |
One-Dimensional Acceleration
Utility of the One-Dimensional Case
As with all vector equations, the equations of kinematics are usually approached by separation into components. In this fashion, the equations become three simultaneous one-dimensional equations. Thus, the consideration of motion in one dimension with acceleration can be generalized to the three-dimensional case.
Useful Digrammatic Representations
Several diagrammatic representations are commonly used to represent accelerated motion.
Deceleration
In physics, the term acceleration denotes a vector, as does velocity. When the acceleration of an object points in the same direction as its velocity, the object speeds up. When the acceleration of an object points in the direction opposite the object's velocity, the object slows down. In everyday speech, we would call the first case "acceleration" and the second case "deceleration". In physics, both cases represent acceleration, but with a different relationship to the velocity.
Constant Acceleration
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Integration with Respect to Time
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If acceleration is constant, the definition of acceleration can be integrated:
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\begin{large}\[ \int_{v_{\rm i}}^{v} dv = \int_{t_{\rm i}}^{t} a\: dt \] \end{large} |
For the special case of constant acceleration, the integral yields:
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\begin{large} \[ v - v_{\rm i} = a(t-t_{\rm i}) \] \end{large} |
which is equivalent to:
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\begin{large} \[ v = v_{\rm i} + a (t-t_{\rm i}) \] \end{large} |
We can now substitute into this equation the definition of velocity,
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\begin{large}\[ v = \frac{dx}{dt}\]\end{large} |
which gives:
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\begin{large} \[ \frac{dx}{dt} = v_{\rm i} + a t - a t_{\rm i} \] \end{large} |
We can now integrate again:
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\begin{large} \[ \int_{x_{\rm i}}^{x} dx = \int_{t_{\rm i}}^{t} \left( v_{\rm i} - at_{\rm i} + a t\right)\:dt \] \end{large} |
to find:
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\begin{large} \[ x - x_{\rm i} = v_{\rm i} (t-t_{\rm i}) - a t_{\rm i} (t-t_{\rm i}) + \frac{1}{2}a( t^{2} - t_{\rm i}^{2}) \] \end{large} |
We finish up with some algebra:
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\begin{large}\[ x = x_{\rm i} + v_{\rm i} (t-t_{\rm i}) + \frac{1}{2} a (t^{2} - 2 t t_{\rm i} + t_{\rm i}^{2}) \] \end{large} |
which is equivalent to:
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\begin{large} \[ x = x_{\rm i} + v_{\rm i} (t-t_{\rm i}) + \frac{1}{2} a (t - t_{\rm i})^{2} \] \end{large} |
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Integration with Respect to Position
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The definition of acceleration can also be integrated with respect to position, if we use a calculus trick that relies on the chain rule. Returning to the definition of acceleration:
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\begin{large}\[ \frac{dv}{dt} = a \] \end{large} |
we would like to find an expression for v as a function of x instead of t. One way to achieve this is to use the chain rule to write:
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\begin{large} \[ \frac{dv}{dx}\frac{dx}{dt} = a \] \end{large} |
We can now elminate t from this expression by using the defnition of velocity to recognize that dx/dt = v. Thus:
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\begin{large} \[ \frac{dv}{dx}v = a \] \end{large} |
which is easily integrated for the case of constant acceleration:
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\begin{large} \[ \int_{v_{\rm i}}^{v} v \:dv = \int_{x_{\rm i}}^{x} a \:dx \] \end{large} |
to give:
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\begin{large}\[ v^{2} = v_{\rm i}^{2} + 2 a (x-x_{\rm i}) \] \end{large} |
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The Utility of Constant Acceleration
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Stringing together a series of constant velocity segments is not usually a realistic description of motion, because real objects cannot change their velocity in a discontinuous manner. This drawback does not apply to constant acceleration, however. Objects can have their acceleration changed almost instantaneously. Because of this, it is often reasonable to approximate a complicated motion by separating it into segments of constant acceleration.