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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}{composition-setup} {table:border=1|frame=void|rules=cols|cellpadding=8|cellspacing=0} {tr:valign=top} {td:width=350|bgcolor=#F2F2F2} {live-template:Left Column} {td} {td:bgcolor=#F9F9FF} h1. Acceleration {excerpt}The time rate of change of [velocity].{excerpt} h3. {toggle-cloak:id=mathrep} Mathematical Representation {cloak:id=mathrep} {latex}\begin{large}\[ \vec{a} = \frac{d\vec{v}}{dt} \qquad \]\end{large}{latex} {cloak} h3. {toggle-cloak:id=1da} One-Dimensional Acceleration {cloak:id=1da} h4. {toggle-cloak:id=util} Utility of the One-Dimensional Case {cloak:id=util} 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. {cloak:util} h4. {toggle-cloak:id=graph} Graphical Representations of Acceleration {cloak:id=graph} 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! Examples that use the graphical representations of acceleration are: {contentbylabel:graphical_representation,constant_acceleration|maxResults=50|operator=AND|showSpace=false|excerpt=true} {cloak:graph} h4. {toggle-cloak:id=decel} The Danger of Deceleration {cloak:id=decel} 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: ||positive acceleration\\positive velocity\\"accelerating"||negative acceleration\\negative velocity\\"accelerating"||negative acceleration\\positive velocity\\"decelerating"||positive acceleration\\negative velocity\\"decelerating"|| |!pos1.gif!|!neg1.gif!|!neg2.gif!|!pos2.gif!| 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} {cloak:decel} h4. {toggle-cloak:id=const} Constant Acceleration {cloak:id=const} h5. {toggle-cloak:id=inttime} {color:maroon} Integration with Respect to Time {color} {cloak:id=inttime} If acceleration is constant, the definition of acceleration can be integrated: {latex}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}{latex}
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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} |
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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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{latex} {cloak:inttime} h5. {toggle-cloak:id=intpos} {color:maroon} Integration with Respect to Position {color} {cloak:id=intpos} 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
...
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} {cloak:intpos} {cloak:const} h4. {toggle-cloak:id=motconst} One-Dimensional Motion with Constant Acceleration {cloak:id=motconst} h5. {toggle-cloak:id=5eqs} {color:maroon} Four or Five Useful Equations {color} {cloak:id=5eqs} For a time interval during which the acceleration is constant, the instantaneous acceleration at any time will always be equal to the average acceleration. Thus, by analogy with the definition of [average velocity|velocity], we can write: {latex}\begin{large} \[ a = \langle a\rangle_{t} = \frac{\Delta v}{\Delta t} = \frac{v - v_{\rm i}}{t - t_{\rm i}} \] \end{large}{latex} Taking this equation as a starting point and using the relationship between average velocity and position {latex}\begin{large} \[ \langle v\rangle_{t} = \frac{\Delta x}{\Delta t} = \frac{x - x_{\rm i}}{t- t_{\rm i}} \] \end{large} {latex} {warning}We removed the bar in the acceleration equation above because acceleration is constant. We *cannot* similarly drop the bar in the average velocity equation, because velocity is *not constant* when acceleration is constant, except in the special case that _a_ = 0 (and in that case, you should really use the simpler model for [constant velocity|1-D Motion (Constant Velocity)]. {warning} lets us derive five very important equations. {info}Three of these equations follow directly from the integrations performed in the section above.{info} {panel:title=Five (or Four) Equations for Kinematics with Constant Acceleration} {latex}\begin{large} \[ x = x_{\rm i} + \bar{v}(t-t_{\rm i}) \] \[ \bar{v} = \frac{1}{2}(v+v_{\rm i}) \] \[ v = v_{\rm i} + a(t-t_{\rm i}) \]\[ x = x_{\rm i} + v_{\rm i}(t-t_{\rm i}) + \frac{1}{2} a (t-t_{\rm i})^{2} \] \[ v^{2} = v_{\rm i}^{2} + 2 a (x-x_{\rm i}) \]\end{large}{latex} {panel} {note}Because the first equation is not specific to the case of constant acceleration (it is simply a definition of average velocity) it is combined with the second equation in the summary on the model specification page for [one-dimensional motion with constant acceleration|1-D Motion (Constant Acceleration)]. {note} h5. {toggle-cloak:id=tf} {color:blue}{+}A Training Flight (Some Typical Exercises)+{color} {cloak:id=tf} {include:Training Flight} {cloak:tf} {cloak:5eqs} h5. {toggle-cloak:id=conautil} {color:maroon}The Utility of Constant Acceleration{color} {cloak:id=conautil} 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. For example, you could be coasting along in a car at a constant 60 mph with zero acceleration when suddenly you see traffic stopped ahead. If you slam on the brakes, your car will still be going 60 mph for an instant, then it will drop to 59, 58, 57, etc. Your acceleration, on the other hand, has almost instantly changed from zero to a substantial acceleration directed opposite your motion. You can feel this abrupt change as a passenger as you are forced against your seatbelt. Similarly, when an airplane begins its takeoff run, you can feel yourself suddenly pressed back in your seat as the plane's acceleration changes almost instantaneously from 0 to a significant forward acceleration. Because of this, it is often reasonable to approximate a complicated motion by separating it into segments of constant acceleration. h5. {toggle-cloak:id=ec} {color:blue}{+}An Exercise in Continuity{+}{color} {cloak:id=ec} {include:An Exercise in Continuity} {cloak:ec} h5. {toggle-cloak:id=fpib} {color:blue}{+}Example Problem -- Fan-Powered Ice Boat{+}{color} {cloak:id=fpib} {include:Fan-Powered Ice Boat} {cloak:fpib} {cloak:conautil} h5. {toggle-cloak:id=grav} {color:maroon}Near-Earth Gravity{color} {cloak:id=grav} One of the most important applications of [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)] is describing the motion of objects moving vertically under the influence of [gravity]. For objects near the surface of the earth, gravity creates a constant pull directed essentially straight toward the ground. {info}Because of the earth's enormous size, "near the surface" applies to anywhere that ordinary people will travel. In fact, even the space shuttle in orbit is experiencing an acceleration of around 8.7 m/s ^2^ (only 12% less than the acceleration of gravity at the surface of the earth) toward the earth's center. (The astronauts do not "feel" this acceleration for reasons to be discussed in a later lesson.) {info} Interestingly, the gravitational acceleration produced by the earth is not only constant on objects near its surface, but it is also exactly the same magnitude for all objects if air resistance is negligible. That value is: {panel:title=Magnitude of the Acceleration of Gravity Near the Surface of the Earth} {latex}\begin{large} \[ g = 9.8\:{\rm m/s}^{2}\]\end{large}{latex} {panel} {note}Air resistance is often non-negligible if 99% accuracy is desired. Neglecting air resistance is generally reasonable for ballpark (good to within 10%) estimates, provided that the object is not of an extremely low density (e.g. a beach ball), or of a shape that interacts with the air to produce lift (e.g. feathers, sheets of paper). The reasonableness of ignoring air resistance depends heavily on the distance of the fall. For example, it is an excellent approximation to ignore air resistance on a baseball dropped from a table, a reasonable ballpark estimate to ignore air resistance on a baseball thrown (hard) straight upward, and a terrible estimate to ignore air resistance on a baseball dropped from an airplane 2 miles above the ground. {note} h5. {toggle-cloak:id=kp} {color:blue}{+}Example Problem -- Keys Please{+}{color} {cloak:id=kp} {include:Keys Please} {cloak:kp} {cloak:grav} h5. {toggle-cloak:id=catup} {color:maroon}Catch-Up Problems By Example{color} {cloak:id=catup} h5. {toggle-cloak:id=st} {color:blue}{+}Example Problem -- Speed Trap{+}{color} {cloak:id=st} {include:Speed Trap} {cloak:st} {cloak:catup} {cloak:motconst} {cloak:1da} {td} {tr} {table} {live-template:RELATE license} |
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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.