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Background

I'm new to physics and math. I stopped studying both of them in high-school, and I wish I hadn't. I'm pursuing study in both topics for personal interest. Today, I'm learning about acceleration as a step toward understanding force, and then related concepts (power, work, etc.)

What I Think I Understand

Acceleration is the change of velocity per second. Example: If we begin with a car at rest, that then begins to move forward, and reaches a velocity of 28 m/s over 10 seconds; then acceleration is calculated as (28-0)/10, which yields acceleration equal to 2.8 m/s/s.

Question

Is it understood that acceleration is (or, is sometimes) an average?

My Thoughts on the Question

The following section is to add context to the questions. It's a transcription of my attempt to answer the questions. I welcome comments on what I've written below.

My car doesn't accumulate speed evenly as its speed changes from 0-100 Km/H. So if I applied the above formula to my car's acceleration, I'd really get something (that seems to me like) an -average- rate of acceleration over 10 seconds.

Is it understood that acceleration is (or, is sometimes) an average?

If that's so; and if F=MA; and if the rate at which my car accumulates velocity over a second is greater during its 10th second of velocity accumulation, than during any prior second; and if I calculate force using only the final second of velocity accumulation; and if I also calculate force using all 10 seconds of velocity accumulation: then F=MA will produce one measure of force applicable to the 10th second, and one measure of force applicable to the 10 second period. But for both those measures to be valid, then F=MA must also express an average amount of force over a period.

I think that changing the unit of time one uses in measuring the natural property that has been ascribed the term 'force', would only change the description of its measure, but not its nature nor its degree.

To me, it also seems that whatever measure of velocity-accumulation per-second an object had some time ago, should not affect the degree of force it has now. For instance, if my velocity increased from 0 to 10,000 M/S in the first half of a second, and then remained at 10,000 M/S for the remainder of the second, and someone (say a passenger) asked me, 'what's our acceleration?'. I could tell him 10,000 M/S/S, but to me that answer seems misleading, as our acceleration at that instant is 0 M/S/S^-∞.

Also, if F=MA, then my vehicle, which is travelling at a constant velocity of 10,000 M/S, would have no 'force' (F=M*0).

Unless F=MA is referring to the 'challenge': if a 1000KG moose steps in the way of my 10,000 M/S vehicle (God help him), and if he remains intact (miraculously), and if he then accumulates a given degree of velocity over 1 second, then the force on the moose would be expressed F=1000KG*(Change in velocity during the second of measurement).

Now, if I were an enormous unbreakable-plane travelling in inter-galactic space, and travelling at a constant speed and direction; and if I encountered my clone, who were oriented as I were oriented, and travelling at a speed equal to mine, and travelling in a direction opposite my direction of travel; and if we had both depleted our fuel reserves long-ago, so that our movement was sustained only by momentum; and if we collided: then I suspect both of us would just stop moving - sandwiched together and perfectly stationary.

Thus the force of my clone is (my mass)x(my deceleration), and my force is (my clone's mass)x(my clone's deceleration)

(I used enormous unbreakable-planes to suppress a mental-image of ricocheting or shattering.)

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  • $\begingroup$ Hi Hal, and welcome to Physics Stack Exchange! Questions should be limited to one (or a few very closely related ones) per post here, so I've edited out the extra parts of your question. Feel free to post them separately. $\endgroup$
    – David Z
    Commented Apr 18, 2013 at 2:44
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    $\begingroup$ I had a very similar question when I was in grade 12 physics. It made a lot more sense after I finally took calculus next semester. (They really should make calculus prerequisite.) $\endgroup$
    – Cruncher
    Commented Jan 30, 2014 at 21:45

3 Answers 3

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The acceleration of an object at a given time $t$ is, in fact, defined as the average acceleration of that object for a small interval around that time in the limit that the interval becomes "infinitely small." The mathematical concept of the limit makes the notion of "infinitely small" precise.

Suppose $v(t)$ represents the velocity of the object moving along a straight line at time $t$. If we want to find its acceleration at that time, what we do is to first determine its average acceleration for some interval of time between time $t$ and $t+\Delta t$ where $\Delta t >0$. This is just given by the change in velocity divided by the change in time over that interval; $$ a_\mathrm{average}(t,t+\Delta t) = \frac{v(t+\Delta t) - v(t)}{\Delta t} $$ Next, we imagine taking that interval infinitely small; the result is the instantaneous acceleration at time $t$. This operation is made notationally precise by the following mathematical notation, and the precise mathematical notion can be defined in terms of the "delta-epsilon" definition found in standard calculus and real analysis texts: $$ a(t) = \lim_{\Delta t \to 0} a_\mathrm{average}(t, t+\Delta t). $$ Putting our two equations together gives the definition of instantaneous acceleration: $$ a(t) = \lim_{\Delta t\to 0} \frac{v(t+\Delta t)-v(t)}{\Delta t} $$ The right hand side of this expression is mathematically referred to as the derivative of the velocity function and is often denoted in physics with an overdot; $$ \dot v(t) = \lim_{\Delta t\to 0} \frac{v(t+\Delta t)-v(t)}{\Delta t}. $$ This allows us to restate the definition of acceleration succinctly as $$ a(t) = \dot v(t) $$ Or said in words, acceleration is the time derivative of velocity.

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  • $\begingroup$ I read about limits this morning. This evening I'll work with the formula you explained and see if I can accrue a more precise understanding of acceleration. Nevertheless, your answer clarified a great deal, even with my understanding of limits absent. Thank you! $\endgroup$
    – Hal
    Commented Apr 18, 2013 at 12:50
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    $\begingroup$ @Hal Calculus is very important even in simple problems like these, because it's where our understanding of instantaneous velocity and instantaneous acceleration comes from and is made precise. You'll want to get comfortable with differentiation, integration, and solving differential equations to get a good grasp on mechanics. $\endgroup$
    – Daniel Blay
    Commented Apr 18, 2013 at 21:48
  • $\begingroup$ @Hal Glad to hear it! Hope you have fun exploring calculus! It's the sh@#. $\endgroup$
    – joshphysics
    Commented Apr 19, 2013 at 6:12
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Acceleration can be used as an average of the change in velocity over the change in time. Acceleration becomes the derivative of velocity when velocity changes over extremely small intervals. If you remember the definition of the derivative, the derivative is a quotient as the difference between x-values,in this case t-values, for time, approaches 0. Acceleration is the change in velocity over change in time. So, it is sometimes an average, but sometimes it is not, especially when acceleration is not constant.

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Partly right. By definition, acceleration is the time derivative of velocity and is defined AT time instant $t$. So we take the time derivative of a velocity function then evaluate it AT $t$. However, experimentally (and even computationally) we usually compute acceleration as the change in velocity $\Delta v$ over a small but finite time interval $\Delta t$. In that sense, it's an average over $\Delta t$.

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