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| Excerpt |
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The |
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time |
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rate |
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of |
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change |
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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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velocity.{excerpt} || PageContents || | {toc:style=none|indent=10px} | h2. Mathematical Definition {latex}\begin{large}\[ \vec{a} = \frac{d\vec{v}}{dt} \qquad \mbox{or} \qquad \vec{a}=\frac{\sum \vec{F}}{m} \]\end{large}{latex} ---- h2. |
One-Dimensional
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Acceleration
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Utility
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of
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the
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One-Dimensional
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Case
...
As
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with
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all
...
...
equations,
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the
...
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
...
separation
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into
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components.
...
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,
...
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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h4. Graphical Representations of Acceleration The one-dimensional form of the definition of acceleration: {latex}\begin{large} \[ a = \frac{dv}{dt}\] \end{large}{latex} indicates that acceleration is the slope of a velocity versus time graph. Since velocity is the derivative of position: {latex}\begin{large}\[ v = \frac{dx}{dt}\]\end{large}{latex} we can also write: {latex}\begin{large}\[ a = \frac{d^{2}x}{dt^{2}}\]\end{large}{latex} Thus, we can also see the effects of acceleration in a position versus time graph like that shown here: !accel.gif! Consider the position vs. time graph shown above. If you were to lay a ruler along the curve of the graph at the origin, the ruler would have to be horizontal to follow the curve, indicating zero slope. Thus, the velocity is zero at the origin. As you follow the curve, however, the ruler would have to be held at a steeper and steeper angle (see the lines added in the graph below). The slope grows with time, indicating that the velocity is becoming more and more positive (the speed is increasing). This positive change in velocity indicates a positive acceleration. In calculus terminology, we would say that a graph which is "concave up" or has positive curvature indicates a positive acceleration. !acceltangents.gif! h4. The Danger of Deceleration Acceleration poses a problem for the specialized vocabulary of physics. The other two major kinematical quantities, [velocity] and [position], have a related scalar quantity. For instance, [distance] is a scalar that is related to [displacement]. [Speed|speed] is a scalar that is related to velocity. If we are discussing instantaneous velocity, then speed is the magnitude of velocity. The acceleration can also be discussed in terms of a vector acceleration or simply the magnitude, but for acceleration we have no special term for the magnitude. The vector is called "the acceleration" and the magnitude is "the magnitude of the acceleration". This can result in confusion. This problem is exacerbated by the fact that in everyday language, we often discuss _distance_, _speed_ and _acceleration_. The everyday definitions of distance and speed are basically equivalent to their physics definitions, since we rarely discuss direction of travel in everyday speech and these quantities are scalars in physics (no direction). Unfortunately, in physics, we usually use the term "acceleration" to refer to a vector, while in everyday speech it denotes a magnitude. The difficulties do not end there. Everyday usage _does_ make one concession to the vector nature of motion. When we discuss acceleration in everyday speech, we usually specify whether the object is "accelerating" (speeding up) or "decelerating" (slowing down). Both terms imply a change in velocity, and so in physics we can call either case "accelerating". The physics way of explaining the difference is: ||everyday term || physics equivalent || | acceleration | acceleration and velocity point in the same direction | | deceleration | acceleration points in the direction opposite the velocity | To understand the physics definition, imagine a child on a playground swing. If you want to help the child swing faster, you must push them in the same direction as they are currently moving (so the acceleration of your push is in the same direction as the child's velocity). If you want to help them slow down, you must push them in the direction opposite their current motion (so that the acceleration of the push points opposite to the velocity). The difference between acceleration and deceleration (in the everyday sense) can also be illustrated graphically: {table}{tr}{th:align=center|bgcolor=#F2F2F2}positive acceleration\\positive velocity\\"accelerating"{th}{th:align=center|bgcolor=#F2F2F2}negative acceleration\\negative velocity\\"accelerating"{th}{th:align=center|bgcolor=#F2F2F2}negative acceleration\\positive velocity\\"decelerating"{th}{th:align=center|bgcolor=#F2F2F2}positive acceleration\\negative velocity\\"decelerating"{th}{tr}{tr}{td}!pos1.gif!{td}{td}!neg1.gif!{td}{td}!neg2.gif!{td}{td}!pos2.gif!{td}{tr}{table} Both the graphs that show "acceleration" have slopes that are steepening with time. The only difference is that one of the graphs has a steepening _positive_ slope and the other has a steepening _negative slope. Both graphs showing "deceleration" have slopes that are approaching zero as time evolves. (Again, one has a negative slope and one has a positive slope.) {warning}It is a very common misconception that a negative acceleration *always* slows down the object it acts upon. This is *not true*. It is important to note that a graph which has a negative slope approaching zero (slowing down) implies a positive acceleration, and a graph which has a negative slope that is steepening (speeding up) implies a negative acceleration. It may help you to remember that the concavity of the graph specifies the direction of the acceleration.{warning} h4. Constant Acceleration h5. Integration with Respect to Time 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:
| Latex |
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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:
| Latex |
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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,
| Latex |
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}\begin{large}\[ v = \frac{dx}{dt}\]\end{large}{latex} |
which
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gives:
| Latex |
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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:
| Latex |
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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:
| Latex |
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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:
| Latex |
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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:
| Latex |
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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} h5. Integration with Respect to Position The definition of acceleration can also |
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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:
| Latex |
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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:
| Latex |
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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:
| Latex |
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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:
| Latex |
|---|
}\begin{large} \[ \int_{v_{\rm i}}^{v} v \:dv = \int_{x_{\rm i}}^{x} a \:dx \] \end{large}{latex} |
to
...
give:
| Latex |
|---|
}\begin{large}\[ v^{2} = v_{\rm i}^{2} + 2 a (x-x_{\rm i}) \] \end{large}{latex} ---- h2. Relevant Examples {contentbylabel:1d_motion} |
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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.