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This is what is written in Resnick Halliday Krane Physics Vol.1

The speedometer of a car indicates it's speed, not it's velocity, because it does not specify a direction. Speed is a scalar, since it lacks directional information. We can also define the average speed as; average speed $=$ $\frac {total \ distance \ travelled} {elapsed \ time}$

Let's say my speedometer shows $\frac{18x} {5}$ km/h. I step on the accelerator for one second and now it shows $\frac {18x} {5}+5$ km/h. So does that mean I travelled $5$ kilometres more in $\frac {1} {3600}$ hour?

But for now leave that question be and there are several more Questions regarding the speedometers.(I don't know anything about cars. My dad might know because he drives one but I don't want to question him for now.)

Since I was small, I frequently used to see these guysScooter roaming all round in the streets. When I grew up, I got the chance to ride these scooters and the first thing that amazed me was the speedometer. To start a scooter from rest you need to turn the accelerator anti clockwise just like shown in the picture.Accelerator of a Scooter. That will start the scooter from rest and will bring it to the speed you have set by rotating the accelerator.

So let's say I rotated the accelerator by a few degrees only and my speedometer shows $20$km/h. To keep the speed constant I would have to keep the accelerator constant at the angle I had rotated earlier to reach $20$km/h. And in order to increase speed I would have to turn the accelerator by some degrees more again in the anticlockwise direction. To deaccelerate it, I would have to turn the accelerator clockwise by some degrees and to completely stop the scooter, I would leave the accelerator back to it's original position(the point where the scooter was at rest)

Now the question is, all those things I have said just earlier, is that change in velocity($\Delta v$) or change in acceleration($\Delta a$) or change in speed($\Delta s$)?

My assumption is since the speed or velocity depends on the change in accelerator or change in acceleration, the change in movement we see while riding a scooter is change in velocity but we always say check the 'speed'. I think that should be Velocity since it depends on the direction.

It's either that or I have a horrible understanding of speed, velocity and acceleration.

What do you guys think?

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    $\begingroup$ "Accelerator" is a bad name for that control on your scooter. It controls the rearward force that the tire applies to the road. If you rode in an ideal world, on a perfectly smooth, level road, with tires that offer no rolling resistance, with no air to cause aerodynamic drag, then that rearward force would be directly proportional to your acceleration. (Issac Newton said, $F=ma$.) But the real world is more complicated. You need a certain amount of force just to keep the bike rolling along at a constant speed (no acceleration.) And, you need more force to go at a faster constant speed. $\endgroup$ Commented Sep 22, 2022 at 14:46

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Why is there an x in the speedometer reading in the first portion?

A 'speedometer' shows your speed, and if you consider you are moving in whatever the 'forward' direction is, that gives you a velocity. If after you press the accelerator for some time the speedometer shows a new value, you have accelerated. What you have gained directly by accelerating is velocity (meters per second in the forward direction), not distance (meters) - you have also gained distance but you don't see this on the speedometer, you get this by integrating your acceleration to get your velocity at each point in time and then integrating velocity from start to end times to get distance moved.

The way your scooter works, it seems as if the dial on the 'accelerator' shows the speed you would like to go at, and then it revs up to that speed and then maintains it. In that case the dial is a speedometer. If instead it showed acceleration, the scooter would continue to accelerate as long as the dial is not at 'zero', and will not maintain a constant speed otherwise.

In short, as vectors, acceleration is the rate of change of velocity, and velocity is the rate of change of the position vector of your object in some coordinate system. If the position changes, then the object moves in a certain direction; the speed plus direction gives the velocity. If the direction in which it is moving changes, or the speed at which it is moving changes, that is an acceleration (or deceleration). The direction in which the direction changes (I know), or the direction in which the speed changes (e.g. decreasing or increasing in relation to the original direction) is the direction of the acceleration, and the amount by which it changes is the magnitude.

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  • $\begingroup$ I understood the speedometer part. My electric scooter works like this It accelerates as long as I have the accelerator turned anticlockwise. When I have my accelerator on full throttle it shows a constant of 24km/h, but when I deaccelerate it it speed remains the same until after some time it's value changes to match the current speed of my scooter. So is that a speedometer? If it is then I have understood what a speedometer is. $\endgroup$
    – AshCAD
    Commented Sep 22, 2022 at 3:45
  • $\begingroup$ I don't understand you scooter. Try this though: en.wikipedia.org/wiki/Accelerometer $\endgroup$
    – Basican
    Commented Sep 22, 2022 at 3:48
  • $\begingroup$ Forget the scooter. Say I go at speed 24km/h and then I decrease my speed. My speedometer which was showing 24km/h now shows 20km/h after 2 seconds. Assuming in that 2 seconds the speed of the vehicle was reduced to 20km/h. Is that how the speedometer works? $\endgroup$
    – AshCAD
    Commented Sep 22, 2022 at 3:53
  • $\begingroup$ Yes, you decreased your speed and the speedometer shows the new reading. The acceleration was -4 km per hour, over two seconds, in the forward direction. You could write that as -2 km/h/s, on average, and then convert the hours to seconds or vice versa to get one coherent unit of time. The point is, the unit of acceleration is distance * time ^(-2). $\endgroup$
    – Basican
    Commented Sep 22, 2022 at 3:58
  • $\begingroup$ Great! I understood the entire thing. That solves the problem $\endgroup$
    – AshCAD
    Commented Sep 22, 2022 at 4:01

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