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| Composition Setup |
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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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}{composition-setup} {excerpt}The time rate of change of [velocity] of an object, or alternately the net [force] on the object divided by the object's [mass].{excerpt} h4. Mathematical Representation {latex}\begin{large}\[ \vec{a} = \frac{d\vec{v}}{dt} \qquad \mbox{or} \qquad \vec{a}=\frac{\sum \vec{F}}{m} \]\end{large}{latex} h4. |
One-Dimensional
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Acceleration
Utility of the One-Dimensional
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Case
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As
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with
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all
...
...
equations,
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the
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equations
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of
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kinematics
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are
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usually
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approached
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by
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separation
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into
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components.
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In
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this
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fashion,
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the
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equations
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become
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three
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simultaneous
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one-dimensional
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equations.
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Thus,
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the
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consideration
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of
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motion
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in
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one
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dimension
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with
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acceleration
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can
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be
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generalized
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to
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the
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three-dimensional
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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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h5. Useful Digrammatic Representations Several diagrammatic representations are commonly used to represent accelerated motion. * [*Position vs. Time Graph*|position versus time graph] * [*Velocity vs. Time Graph*|velocity versus time graph] * [*Motion Diagram*|motion diagram] h5. 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]. h5. Constant Acceleration h6. Integration with Respect to Time If acceleration is constant, the definition of acceleration can be integrated: {latex}\begin{large}\[ \int_{v_{\rm i}}^{v} dv = \int_{t_{\rm i}}^{t} a\: dt \] \end{large}{latex} |
For
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the
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special
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case
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of
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constant
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acceleration,
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the
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integral
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yields:
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}\begin{large} \[ v - v_{\rm i} = a(t-t_{\rm i}) \] \end{large}{latex} |
which
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is
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equivalent
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to:
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}\begin{large} \[ v = v_{\rm i} + a (t-t_{\rm i}) \] \end{large}{latex} |
We
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can
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now
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substitute
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into
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this
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equation
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the
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definition
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of
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velocity,
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}\begin{large}\[ v = \frac{dx}{dt}\]\end{large}{latex} |
which
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gives:
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} \begin{large} \[ \frac{dx}{dt} = v_{\rm i} + a t - a t_{\rm i} \] \end{large}{latex} |
We
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can
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now
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integrate
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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}{latex} |
to
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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}{latex} |
We
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finish
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up
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with
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some
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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}{latex} |
which
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is
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equivalent
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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}{latex} h6. Integration with Respect to Position |
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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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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: {latex}\begin{large}\[ \frac{dv}{dt} = a \] \end{large}{latex} |
we
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would
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like
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to
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find
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an
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expression
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for
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v
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as
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a
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function
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of
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x
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instead
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of
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t
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.
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One
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way
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to
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achieve
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this
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is
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to
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use
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the
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chain
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rule
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to
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write:
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}\begin{large} \[ \frac{dv}{dx}\frac{dx}{dt} = a \] \end{large}{latex} |
We
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can
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now
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elminate
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t
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from
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this
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expression
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by
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using
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the
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defnition
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of
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velocity
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to
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recognize
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that
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dx
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/
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dt
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=
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v
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.
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Thus:
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}\begin{large} \[ \frac{dv}{dx}v = a \] \end{large}{latex} |
which
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is
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easily
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integrated
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for
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the
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case
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of
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constant
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acceleration:
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}\begin{large} \[ \int_{v_{\rm i}}^{v} v \:dv = \int_{x_{\rm i}}^{x} a \:dx \] \end{large}{latex} |
to
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give:
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}\begin{large}\[ v^{2} = v_{\rm i}^{2} + 2 a (x-x_{\rm i}) \] \end{large}{latex} h6. The Utility of Constant Acceleration 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. |
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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.