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Here we have a vehicle, that accelerates, decelerates suddenly, gains a little more speed and then decelerates until it stops.

enter image description here

[The x axis is for the time in seconds!, t(s) the photo didn't encompass it]

There are two moments that drew my attention: '$a$ to $b$' or '$c$ to $d$'

In 'a to b', we have a very sudden acceleration $a$ (which is working against the speed) that costs our vehicle 10m/s in only 0.5 second. Which gives a acceleration ($a$) of 20m/s^2. Here we have the greatest modulus for $a$ (acceleration) in the entire trajectory.

On the other hand, during 'c to d', we have a much more modest $a$ (acceleration), also working against the speed. But it'll work longer resulting in a greater ∆v (V-V0), the vehicle loses a total of 30m/s and stops, but it takes 5,5 seconds to do so.

I'd like to know, by definition, which would be the greatest deceleration:

The first momment ('a to b') in which we have a more sudden change in speed ($a$ has the greatest modulus working against speed), but it doesn't result in a loss of speed as great as the second: 'only' 10m/s

or

The second case ('c to d'), which encompasses the greatest loss in speed: 30m/s (v-v0), eventually stopping the car but taking more time to do so.

In other words, which defines the greatest deceleration in an interval of time? The relative deceleration, caused by how fast it makes the vehicle lose speed. Or an absolute deceleration, caused by how much speed one vehicle loses regardless of time.

In other words, is deceleration to be interpreted as the dictionary suggests:

Oxford Dictionary: The reduction in speed or reduction in rate. Speed: Speed brakes enable the aircraft to carry out rapid deceleration Rate: a deceleration in econimic growth

Or is deceleration interpreted as the rate of reduction in speed.

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Acceleration is defined by:

$$a = \frac{\mathrm{d}v}{\mathrm{d}t}$$

This means that the rate of change of velocity is the acceleration. If we continue this train of thought, the greater the rate of change of velocity, the greater the acceleration.

Graphically, rate of change is the slope, which means the greater the slope, the greater the rate of change. You are right is saying that in the time period you've labelled "a", the vehicle experiences the greatest acceleration. You might be hesitant to say that because the acceleration is opposite of the velocity, but regardless, the acceleration is the greatest. If we look at the graph, the slope is the steepest, even though it is negative, which means the rate of change is the greatest meaning acceleration there is the largest.

The reason why period "b" doesn't have the greatest acceleration is because the slope is not steeper than period "a". It doesn't matter how long period "b" is compared to "a", the fact is that the slope is what determines the acceleration, not the length of time.

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  • $\begingroup$ Thank you for such an elaborate answer. I'm deeply sorry, but what I wanted to say on the title was "Which is the greater deceleration". What is the concept of greater deceleration, a greater change in speed, regardless of time needed to do so, or a more sudden change of speed (that, if kept longer would result in drastic speed change), but because it lasts less in the situation it doesn't interfere with the speed as much in absolute values? $\endgroup$ – Matheus Bezerra Soares Feb 6 at 20:36
  • $\begingroup$ @MatheusBezerraSoares The answer is given by the definition in the first sentence. Deceleration is an acceleration vector which is the opposite direction of the velocity vector. The acceleration is the slope, not the $\Delta v$. $\endgroup$ – Bill N Feb 6 at 21:16
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    $\begingroup$ @MatheusBezerraSoares Deceleration means to accelerate in a direction opposite of that of the velocity. If the vehicle had a velocity in the $+\hat{x}$ and an acceleration in the $-\hat{x}$, we would call that deceleration. Deceleration is acceleration. This would not change the answer though, as period "a" would still have the greatest acceleration and deceleration. $\endgroup$ – MrMineHeads Feb 6 at 21:19
  • $\begingroup$ @Bill N, I'm trying to understand that definition. By the dictionary deceleration means: the reduction in speed. Does it mean, in two different intervals of time, which one had the greatest deceleration: the one who had a very sudden $a$, or the one more lengthy that resulted in a greater loss of speed? I've edited my question again, I hope it's more clear. Please read it again if you can spare the time to do so. And I'm sorry for not being able to pose the question really clear, as I lack the mathematical terminologies. $\endgroup$ – Matheus Bezerra Soares Feb 7 at 0:08
  • $\begingroup$ I've edited my Question once again. I hope it's clearer now. Please read it, if you can spare the time. I don't think I could explain very well on comments. $\endgroup$ – Matheus Bezerra Soares Feb 7 at 0:16
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Acceleration is

$$a=\frac{dv}{dt}$$

So the magnitude of the acceleration in your graph is greatest where the slope is the greatest (+ or -). In this case, maximum is the negative slope of the segment following $a$ (max deceleration). The maximum positive acceleration is the first line segment.

UPDATE:

Your edits indicate that by "bigger acceleration" you meant the greater the change in velocity. But acceleration is the rate of change in velocity (positive or negative) not the amount of change in velocity. This is at the root of the confusion.

That said, the magnitude of the change in velocity does relate to an important difference between intervals a-b and c-d. That difference is the change (reduction) in kinetic energy of the vehicle in the intervals, which in turn is related to the work done on the vehicle to slow the vehicle down for the two intervals.

The work-energy theorem states that the net work done on an object equals its change in kinetic energy, or

$$W_{net}=Fd=\frac{mv_{f}^2}{2}-\frac{mv_{i}^2}{2}$$

Where $v_f$ and $v_i$ are the final and initial velocities, $d$ is the stopping distance for the two intervals, $d_{ab}$ and $d_{cd}$, and $F$ is magnitude of the constant stopping force equal to $ma_{ab}$ and $ma_{cd}$ for the two intervals.

From the velocities on the graph, clearly the magnitude of the work done (by the brakes, engine, etc.) slowing the vehicle down in interval c-d is greater than in interval a-b. In both cases the net work done is negative, which simply means the work takes kinetic energy away from the vehicle.

Hope this helps.

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  • $\begingroup$ Thank you for your reply. I'm deeply sorry because the title was wrong, I wanted to know which of the present concepts would result in a 'bigger deceleration': 1) A greater lost of speed that takes more time. 2) A more sudden loss of speed (with greater acceleration 'a'), but that would last shorter. $\endgroup$ – Matheus Bezerra Soares Feb 6 at 20:41
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    $\begingroup$ @MatheusBezerraSoares Sorry but I'm still not sure what you want. The maximum deceleration (negative rate of change in velocity) occurs after $a$. What exactly do you mean by "bigger deceleration"? $\endgroup$ – Bob D Feb 6 at 22:18
  • $\begingroup$ Thank you for your patience. I've Edited my question's title, image and text. I hope it's clearer now. Please read it if you can spare the time. As I don't think i've got enough characters on a comment to explain here. $\endgroup$ – Matheus Bezerra Soares Feb 7 at 0:16
  • $\begingroup$ @MatheusBezerraSoares See my update based on your edits. $\endgroup$ – Bob D Feb 7 at 11:37
  • $\begingroup$ Thank you so very much. I see now that the work done to the object was much greater, because it resulted in a bigger change in speed. In my Oxford dictionary we have the concepts of acceleration: rate of change of speed. And deceleration: reduction of speed or rate (deceleration in economic growth) I was already accustomed with the concepts of 'negative acceleration' and retarded movemnt. But I went for the definition of the dictionary of deceleration: if deceleration = reduction in speed, once the reduction was the greatest in c-d, then we would have the greatest deceleration. $\endgroup$ – Matheus Bezerra Soares Feb 7 at 13:19

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