lncognito

Actually, the speed changes constantly as well. It just won't go beneath zero.
I realized halfway through writing this that this might actually have been what you meant, but meh... better to clarify it for confused TS (who really should have figured it out by now)

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magnus_maximus

Nope.


So close, so close.

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whoomit

Without the 't'!?

magnus_maximus

Nope. Different altogether. None shall discover my true identity!

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MAC2322

Forgive the red marker, I'm all out of other colors.

Anyway, as you can see from the graphs, Velocity is the only quantity whose sign changes. I avoided using dx/dt and dv/dt for velocity and acceleration respectively since I doubt your Physics class is Calculus-based, but for those of you who do know Calc I, derivatives are what I was approximating with my Δy/Δt and Δv/Δt.

I also marked the time t1 at which the object reaches its highest point so you could see how different quantities are changing (or not changing) at that time.

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archenemyfan

That was awesome. Thank you for taking your time to do that, I appreciate it.

Also my physics class is calc based.

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entity0009

For those of you doubting that gravity is the only acceleration, whatever force that caused the ball (or whatever, I can't remember) to start its trajectory happened BEFORE t=0 and isn't a subject of study for the problem. A lot of people get confused when a is working in an opposite direction to v.

MAC2322

No problem. If you have any other questions, feel free to ask (although I may not get back to you right away since I have to go to class myself).

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Arby911

Ok, not a physics expert here, so could someone explain to me (using short words and small sentences) why acceleration isn't changing sign?

Seems to me when the object leaves the starting point it's accelerating at -9.8 m/s^2 (or as the rest of us call that, decelerating). It then reaches apogee where it's at 0 m/s, and begins on its downward path at an acceleration of +9.8 m/s^2

?

As I've stated, this isn't my area of expertise, and while I have some small understanding of it, I'll not pretend to be an expert, so be gentle.

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ChaosInside

Acceleration is always negative in this case. Initially it has an upward velocity because of the negative acceleration it will slow down until it is 0 m/s. At that point the negative accceleration will cause the velocity to go into the minus and accelerate it in the negative direction if you will.

To make it a little easier to understand maybe, flip the axis so the acceleration is positive and the particle travels in the +y direction. In that case the velocity would get more positive with time. This works exactly the same, but in the negative y direction instead of the positive.

Did this clear things up a bit?

Also noteworthy: Velocity is a vector quantity. Speed is the norm of that vector and is is a scalar that is always greater than or equal to 0. This means that the speed is the square root of the sum of the squared components of the velocity vector. This is the same as the square root of the dot product of the velocity vector with itself, for those with a mathematical background.

Edit: Clarification and typo's.

J-Dawg158

Acceleration is a vector quantity( meaning it can be broken down into componants like left and right, or up and down, in other words, you can describe any 2 dimensional motion in terms of so far to the left/right and so far up/down.)

In physical terms, a change in sign means a change in direction. If the acceleration changed on the trip back down that would imply the now gravity is pulling you up instead of down.

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Arby911

Would it be correct to say that what I was describing above was the effect of acceleration, but not acceleration itself, and thus my error, or am I still not seeing it?

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ChaosInside

Well, you were decently correct though. After it reaches 0 m/s it will accelerate downwards at 9.81 m/s^2. But since it is going in the -y direction it will accelerate at -9.81 m/s^2.

J-Dawg158

If you refer to it's effect on velocity, sure, but it's a little cumbersome. Essientially what you're trying to say is when velocity is positive and acceleration is negative then you decelerate (velocity is decreasing due to opposite directions), but when velocity and acceleration are both negative, it accelerates (velocity is increasing, but in negative direction.)

It's that whole rule of a negative times a negative equals a positive that causes this, but when it comes to physics if you don't pay attention to the direction(signs) then you open yourself up to many careless errors like inadvertantly adding two things that should have been subtracted.

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Arby911

Great, thanks.

I've actually learned something useful in the Pit...not sure if I'm amazed or disappointed...

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J-Dawg158

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Arby911

I wonder if that 'The More You Know' campaign is the longest running PSA?

I know it's been around for a very long time, probably longer than most of the residents here...

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MAC2322

Perhaps a better way to think about it is in terms of direction rather than positive and negative. Rather than thinking of negative acceleration as 'slowing down', we think of it as 'accelerating downwards'. Going by this definition, if you're traveling upwards but accelerating downwards, then yes, your speed is going to decrease, but once you turn around, you're still accelerating downwards, so you speed up.

The speed graph I drew earlier reflects this, with the speed going down and then back up as it changes direction. The velocity graph actually shows the exact same thing, only that, after t1, it's going higher in the negative direction. I suppose the final idea is that the negative sign is an indicator of direction rather than actually meaning 'less than 0'.

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oneblackened

not QUITE right. Velocity is speed in a direction and acceleration is the rate at which velocity changes. This is why something going in a circle is always accelerating even it's moving at the same speed.

OP: Acceleration.

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