#1

A ball is thrown upward from y=0 and then falls back to its starting poimt. Let up be Positiive. which of the following changes sign during the flight?

Velocity

Acceleration

Y

None

So, my meagre knowledge of physics tells me that acceleration in this case is always negative, because it's gravity. But isn't position + during the flight up and - during the flight down?

And isn't velocity the same? I can only choose one answer, I need halp!.

Velocity

Acceleration

Y

None

So, my meagre knowledge of physics tells me that acceleration in this case is always negative, because it's gravity. But isn't position + during the flight up and - during the flight down?

And isn't velocity the same? I can only choose one answer, I need halp!.

#2

Velocity is the speed; acceleration is the rate at which is speeds up or down.

#3

Acceleration duh

#4

Acceleration, you're right.

#5

If I kick a ball straight up at 20 m/s, will it stay at 20m/s the entire journey?

#6

Velocity duh

#7

Acceleration duh

So I'm confuzzled. Does acceleration change from positive to negative? I thought that gravity was always Negative.

Also I should've phrased it better when I said " isnt velocity the same". I meant doesn't velocity change from positive to negative also?

#8

Velocity, assuming I'm understanding right, once the ball reaches it's apex and starts falling it will have negative velocity in terms of your grid, whereas acceleration will be a constant -

I'm also operating under the assumption that the velocity and acceleration hitting 0 doesn't count.

On the other hand it could be acceleration since depending on our frame of reference the acceleration becomes positive on the way down since it's adding to velocity instead of subtracting, actually that's probably what they are looking for. This is why my physics professor insisted on labeling opposing forces in degrees instead of dealing with positive and negative.

*g*assuming you only track the ball once it's finished accelerating upwards, and Y would only be negative if the ball ended up at a lower point than where it started.I'm also operating under the assumption that the velocity and acceleration hitting 0 doesn't count.

On the other hand it could be acceleration since depending on our frame of reference the acceleration becomes positive on the way down since it's adding to velocity instead of subtracting, actually that's probably what they are looking for. This is why my physics professor insisted on labeling opposing forces in degrees instead of dealing with positive and negative.

*Last edited by Kid_Thorazine at Oct 4, 2012,*

#9

answer is velocity

Acceleration remains constant (-g)

velocity starts of positive (as the ball goes upwards) and slowly decreases till a point in midair where for a very short moment velocity = 0 (when it is at it's maximum height)

the ball then starts falling back down with negative velocity (because it is heading downwards)

y does not change sign because negative would imply it ended up lower than the original position

Acceleration remains constant (-g)

velocity starts of positive (as the ball goes upwards) and slowly decreases till a point in midair where for a very short moment velocity = 0 (when it is at it's maximum height)

the ball then starts falling back down with negative velocity (because it is heading downwards)

y does not change sign because negative would imply it ended up lower than the original position

*Last edited by xyber56 at Oct 4, 2012,*

#10

Velocity will change constantly during the flight, but as long as it returns to its starting point it was end up having zero velocity. Acceleration will always be -9.8ms-2 down, as there is no longer any force exerted on the ball after it has left the launching device (your hand, I presume)

#11

The answer is velocity because it is a vector quantity, hence speed and direction are taken into account. The direction that the ball is travelling in changes throughout the flight, so on the way back down the velocity becomes negative.

#12

I'm also operating under the assumption that the velocity and acceleration hitting 0 doesn't count.

On the other hand it could be acceleration since depending on our frame of reference the acceleration becomes positive on the way down since it's adding to velocity instead of subtracting, actually that's probably what they are looking for. This is why my physics professor insisted on labeling opposing forces in degrees instead of dealing with positive and negative.

But if velocity is negative and acceleration is also negative doesn't it add to the velocity because they're both in the same direction?

And thanks for the replies , i'm convinced that Velocity is indeed the right answer.

#13

Using a simple model for this

s=u*t +0.5*a*(t^2)

where s is displacement

u is initial velocity

t is time

a is acceleration

The only force involved is the force due to gravity (constant) at 9.81ms^-2 in the negative direction

Differentiating the previous equation gives an equation for velocity

v=u+a*t

Clearly at t=0, v=u, i.e. the velocity is equal to the initial positive value

If you plotted this equation then you would find a negative slope starting at v=u, t=0. Then crossing the axis at v=0, t=u/a i.e the point at which the ball is stationary. it would then become negative as the ball accelerates towards the ground.

tl;dr acceleration is constant, displacement (y) it always positive, velocity goes from positive to negative.

s=u*t +0.5*a*(t^2)

where s is displacement

u is initial velocity

t is time

a is acceleration

The only force involved is the force due to gravity (constant) at 9.81ms^-2 in the negative direction

Differentiating the previous equation gives an equation for velocity

v=u+a*t

Clearly at t=0, v=u, i.e. the velocity is equal to the initial positive value

If you plotted this equation then you would find a negative slope starting at v=u, t=0. Then crossing the axis at v=0, t=u/a i.e the point at which the ball is stationary. it would then become negative as the ball accelerates towards the ground.

tl;dr acceleration is constant, displacement (y) it always positive, velocity goes from positive to negative.

#14

Velocity references speed and direction. Acceleration is a change in the rate of speed, which is always positive.

*Last edited by JagerSlushy at Oct 4, 2012,*

#15

You are wrong.

shit, I was hoping nobody saw it before I did. but yeah, I caught that right after I posted it

leave it to a mechanical engineer to fuck up a physics question...

*Last edited by JagerSlushy at Oct 4, 2012,*

#16

But wouldn't it need a positive acceleration to start with? Seeming as the ball is thrown?

Velocity definitely changes though.

Velocity definitely changes though.

#17

The answer is velocity because it is avector quantity,hence speed and direction are taken into account. The direction that the ball is travelling in changes throughout the flight, so on the way back down the velocity becomes negative.

I was waiting for someone to say it.

#18

Velocity, assuming I'm understanding right, once the ball reaches it's apex and starts falling it will have negative velocity in terms of your grid, whereas acceleration will be a constant -gassuming you only track the ball once it's finished accelerating upwards, and Y would only be negative if the ball ended up at a lower point than where it started.

I'm also operating under the assumption that the velocity and acceleration hitting 0 doesn't count.

On the other hand it could be acceleration since depending on our frame of reference the acceleration becomes positive on the way down since it's adding to velocity instead of subtracting, actually that's probably what they are looking for. This is why my physics professor insisted on labeling opposing forces in degrees instead of dealing with positive and negative.

Your answer did not need to be this long.

#19

I was waiting for someone to say it.

isn't acceleration a vector quantity aswell?

#20

isn't acceleration a vector quantity aswell?

Yes, but in this model where a ball is thrown up, the acceleration is due to the force of gravity, which is assumed to be constant and always negative (as the ball is always attacted back to the ground)

#21

Velocity is speed in a certain direction. Seeings as the direction changes, i'd say that's your answer.

#22

Your answer did not need to be this long.

It wasn't initially, half of it is me second guessing my first answer.

#23

ill give you an easy trick for you to use whenever you encounter such a problem:

whenever there's change in movement from the original direction to the opposed direction, velocity changes sign, and at the point where the change of directions happened, is whete the velocity is equal to zero.

other info, acceleration is always depending on an applied force, in your case gravity, and if there is no other forces applied during the movement, acceleration will stay constant, so in you case it will stay constant.

whenever there's change in movement from the original direction to the opposed direction, velocity changes sign, and at the point where the change of directions happened, is whete the velocity is equal to zero.

other info, acceleration is always depending on an applied force, in your case gravity, and if there is no other forces applied during the movement, acceleration will stay constant, so in you case it will stay constant.

#24

The answer is velocity because it is a vector quantity, hence speed and direction are taken into account. The direction that the ball is travelling in changes throughout the flight, so on the way back down the velocity becomes negative.

This. In much better words than I could put it in.

Although I'm still not convinced that gravity and accelleration are the exact same thing and not just correlated.

However, I'm a Communications major and haven't had to do this kind of shit in the greater half of a decade.

#25

But wouldn't it need a positive acceleration to start with? Seeming as the ball is thrown?

Velocity definitely changes though.

These problems always start a the time when the ball or whatever has left whatever is producing a force on it. So, what is happening is that something (persons arm) is producing a force on it. Once it has left the arm, it is now in free 'fall' motion. There is nothing to accelerate it upward anymore, only gravity to pull it down. The higher the velocity the longer it will keep moving in the upward direction. So, you would most likely have an acceleration that gives the object speed but think of it as a force that creates acceleration F = ma.

Although I'm still not convinced that gravity and accelleration are the exact same thing and not just correlated.

Gravity is a Force that causes acceleration on objects within it's field. F due to gravity = (G)m1m2/d^2.

G is universal gravitation constant, m1 and m2 are masses of two objects, and those three terms divided by the distance between them squared gives the force of gravity. Which you can then apply to F = ma (I'm pretty sure... it's been awhile since I've done this).

*Last edited by ChrisBW at Oct 4, 2012,*

#26

Yeah, it's velocity.

#27

Velocity. I think everyone's covered why it's not acceleration, but not why it's not position.

Your problem states that it starts at Y=0, goes up and comes back to it's starting point at 0 therefore it was always positive and didn't change sign.

Your problem states that it starts at Y=0, goes up and comes back to it's starting point at 0 therefore it was always positive and didn't change sign.

#28

Gravity is just an acceleration. Gravitational force is caused by distortions in space/time.

Magnus, what's your last name?

#29

Seximus Decimus Aurelius.

>_>

<_<

So it doesn't begin with a

*J*?...

>_>

#30

Do you have....

Ginger hair? >_>

#31

acceleration stays at -9.81m/(s^2). velocity is the property that changes, because the direction in which the object is travelling changes. SPEED doesn't change, velocity does. look into the difference if you don't know it already.

#32

No.

GTFO. That's just insulting brah.

Oh, okay. So your surname isn't Johnston(e)?

#33

Actually, the speed changes constantly as well. It just won't go beneath zero.acceleration stays at -9.81m/(s^2). velocity is the property that changes, because the direction in which the object is travelling changes. SPEED doesn't change, velocity does. look into the difference if you don't know it already.

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)

#34

Nope.

So close, so close.

Without the '

*t*'!?

#35

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.

#36

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

Also my physics class is calc based.

Also my physics class is calc based.

#37

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.

#38

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

Also my physics class is calc based.

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).

#39

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

?

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.

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.

*Last edited by Arby911 at Oct 5, 2012,*

#40

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.

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.

*Last edited by ChaosInside at Oct 5, 2012,*