The positive/negative sign dilemma - vectors applied

[Bhootnike]

Howdy all,

Gaat ah physics question for y'all.

When doing simple motion questions, a situation often arises, when do I label something, e.g. 'a' or 'v', or' x'/'d'/'s' to be negative or positive?
Consider the following example:
A rocket is launched vertically and at 100m above ground, when travelling at 20m/s, an object falls from the rocket and drops to land.
Assume acceleration due to gravity is downwards.

So when labelling the variables, we have:
u= 20
a= - 10 
x/d/s= -100
v=? --> going to be nagative
t=?

My question is:
My understanding is that we assign acceleration to be negative since we in physics denote anything going UP as positive, and anything going DOWN as negative.
Since the object is going DOWN, we must assign 'a' as negative.
Is this correct physical understanding ?
Secondly, displacement obviously is a vector quantity, and it tells us how far from original starting point an object is.
In this question, a negative displacement value is assigned.
Why? - my understanding, complementing what I just said about 'a', is that the object itself travelled 100m away from land, then got dropped. Now at the point of release, a is going to be 10m/s^2 DOWN, or -10m/s^2, and displacement, is going to be 100m away UP from the land. So would it not be positive?
edit - the displacement was being calculated from the rocket, not the land.

-I would associate negative displacement to be only under circumstances where the object has gone BELOW the original starting point, for example, the motion of a mole from the surface of earth to 10m down towards the core of the earth. Or if a person fell of a building, then their displacement is going to be, 'x'  m away from the origin, DOWN from origin, and thus, it'll be NEGATIVE, because as mentioned, we associate anything going UP to be positive, likewise, anything going down to be negative.

Can someone please clarify this.

Hope y'all hayve a naice dey
*Giddy up dobbin!*
 

[yawho]

'Since the object is going DOWN, we must assign 'a' as negative.'    Direction of acceleration a has nothing to do with direction of velocity v.

'my understanding, complementing what I just said about 'a', is that the object itself travelled 100m away from land, then got dropped. Now at the point of release, a is going to be 10m/s^2 DOWN, or -10m/s^2, and displacement, is going to be 100m away UP from the land. '  If the object is 100m above its initial position, displacement is +100 m.

[b^3]

In this case though, we need to take the starting position as when the object is released, as we aren't modelling the first 100 m from launch, just that it starts at 100 m above the ground. So we set 100 m above the ground as the starting point, x is going to be where the object is at the time t, so when it hits the ground, it will be 100 below its starting position, i.e. -100m. Also the best way to do it is to stick to up being positive, and remember that gravity will always act downwards, i.e. -10m/s^2. Initial velocity will be up so positive.

So if you wanted time t, then you would have
u=20 m/s
a=-10 m/s^2
x=-100 m
t=?

and solve

[Bhootnike]

'Since the object is going DOWN, we must assign 'a' as negative.'    Direction of acceleration a has nothing to do with direction of velocity v.

'my understanding, complementing what I just said about 'a', is that the object itself travelled 100m away from land, then got dropped. Now at the point of release, a is going to be 10m/s^2 DOWN, or -10m/s^2, and displacement, is going to be 100m away UP from the land. '  If the object is 100m above its initial position, displacement is +100 m.

Yep, I know that, I didn't write that properly - I meant the object is moving in downwards direction so a must be negative since direction has to be taken into account.

And yes,  - from the ground!

I think I just realised something:

When the variables were labelled, I think they may have been labelled as, 'displacement from ROCKET' rather than displacement from land..., so thats why the displacement is negative, because as its been defined that when something moves down, it has negative vector quantities, so thus, in this example, object falls from rocket and will thus have a -ve displacement and -acceleration and -velocity.

^ Is that right?!

[Bhootnike]

In this case though, we need to take the starting position as when the object is released, as we aren't modelling the first 100 m from launch, just that it starts at 100 m above the ground. So we set 100 m above the ground as the starting point, x is going to be where the object is at the time t, so when it hits the ground, it will be 100 below its starting position, i.e. -100m. Also the best way to do it is to stick to up being positive, and remember that gravity will always act downwards, i.e. -10m/s^2. Initial velocity will be up so positive.

So if you wanted time t, then you would have
u=20 m/s
a=-10 m/s^2
x=-100 m
t=?

and solve

^ yay

[Lasercookie]

My understanding is that we assign acceleration to be negative since we in physics denote anything going UP as positive, and anything going DOWN as negative.
Since the object is going DOWN, we must assign 'a' as negative.
Is this correct physical understanding ?

Take note of the definition of acceleration - the rate of change of velocity: - it's dependent on the change in velocity, so if velocity begins to decrease, acceleration will decrease.

What if you had a car driving along a road (no ups and down)? Say the car is currently moving at 40 km/h.

And let's say it's accelerating: If your acceleration is positive, the car is speeding up. (velocity is increasing). If the acceleration is negative, the car is...? Slowing down. (velocity is decreasing).

And that's all that negative acceleration is. You could have other situations such as if you had the rocket being launched, it's moving up at 100 m/s and then something breaks in the rocket and the rocket begins to slow down - in other words, have a negative acceleration - for a period of time until it's velocity decreases to zero and it starts falling the other way, it would still be moving up (and hence have a positive displacement).

And assuming that negative acceleration stays put, it's velocity will soon become negative (the direction changes, as signified by the negative) and then the rocket would start falling to the ground. (but yeah, what I was trying to point out is that positive direction doesn't really mean that acceleration must be positive)

[Bhootnike]

My understanding is that we assign acceleration to be negative since we in physics denote anything going UP as positive, and anything going DOWN as negative.
Since the object is going DOWN, we must assign 'a' as negative.
Is this correct physical understanding ?

Take note of the definition of acceleration - the rate of change of velocity: - it's dependent on the change in velocity, so if velocity begins to decrease, acceleration will decrease.

What if you had a car driving along a road (no ups and down)? Say the car is currently moving at 40 km/h.

And let's say it's accelerating: If your acceleration is positive, the car is speeding up. (velocity is increasing). If the acceleration is negative, the car is...? Slowing down. (velocity is decreasing).

And that's all that negative acceleration is. You could have other situations such as if you had the rocket being launched, it's moving up at 100 m/s and then something breaks in the rocket and the rocket begins to slow down - in other words, have a negative acceleration - for a period of time until it's velocity decreases to zero and it starts falling the other way, it would still be moving up (and hence have a positive displacement).

And assuming that negative acceleration stays put, it's velocity will soon become negative (the direction changes, as signified by the negative) and then the rocket would start falling to the ground. (but yeah, what I was trying to point out is that positive direction doesn't really mean that acceleration must be positive)

oh nice, thats helpful! especially the last bit! good tip to remember!!

1 more scenario!:
1) firstly, in a question, must we define what we mean by positive/negative?
2) therefore, in the same question we can't change our definition of +ve, -ve yes? because, i have a worked solution here:

A ball is thrown upwards at initial v of 50m/s, from top of 120m building.
a)find time taken to reach max height
b) max height reached
c) time of flight
d) velocity with which ball hits ground

A)u=50, a= -10, x = ? t= ? v= 0 at max height
so v= u + at
thus t= 5 secs

b)t=5. a =-10, u=50, x=? v=0
v^2 = u^2 +2ax
x = 125m from top of building, so total height reached = 245m

c) downwards from highest pt:
u=0, a = -10, x = -245, t= ?
x= ut+1/2 at^2
so t= 7s
thus T = 12 s

d) from initilal pt:
v= ?
u=50
a=-10
x=-120
thus v^2 = u^2 + 2ax
so v= -70
so v = 70m/s down

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If you look at what i've bolded, is my speculation right in that the person has used the definition of up and down in different ways throught this question?

in a, the person has said a = -10 for a ball moving upwards, thus implictly defining the direction UPWARDS as being negative, and vice versa for downwards.
then in b, they've done the same, good.
in C however, they've figured out the time from the highest point to the ground.
so they're looking at the downwards journey.
but, the person has said a = -10, and x = -245m ?
likewise in c, we're looking at velocity at which ball hits land, but the person has said a = -10 and x = -120 ?

has the person used 2 different definitions of up/down ?
or have they generalised by assuming that acceleration due to gravity, is ALWAYS, in any question negative. because the direction of x throughtout the working is consistent, i.e. in a , if it was just assumed that g will always be -10, and that up = positive, and down  = negative, then its consistent. e.g. look at c, theyve said x = -245 m , and its even written on my sheet, 'since its going down ' .
likewise in d, its from starting pt to ground, so downwards direction so x is -ve.

cause im not sure if thats what b^3 meant by :

Also the best way to do it is to stick to up being positive, and remember that gravity will always act downwards, i.e. -10m/s^2. Initial velocity will be up so positive.

[Lasercookie]

cause im not sure if thats what b^3 meant by :

Also the best way to do it is to stick to up being positive, and remember that gravity will always act downwards, i.e. -10m/s^2. Initial velocity will be up so positive.

Okay, I'm not sure what you're asking here. Are you asking if gravity acts downwards?

If so, take a watch of this pretty basic explanation of gravity. It looks like you're overcomplicating things: http://www.youtube.com/watch?v=CUexAhUcx_8

And now onto the other part of your question:


To start off, let's explicitly define negative as downwards, and positive as upwards. This is also the definition that whoever wrote those solutions used, and also the definition that most people use in most cases. I'm never going to change from this (always be consistent with whatever definition you're using, that's the key to it).

I think you've missed the point I was trying to make in my post.

in a, the person has said a = -10 for a ball moving upwards, thus implictly defining the direction UPWARDS as being negative, and vice versa for downwards.

Simply put, no, it doesn't implicitly define the direction upwards as being negative. Just because acceleration is in one direction, does not mean that displacement is in the exact same direction. That's the point that I made earlier.

I think you've forgotten to think in terms of forces. Forces act on objects and they are able to have an affect on objects. They have a linear relationship with acceleration  (). Remember that gravity is a force. 

So, a force is something that has an affect on the motion of the ball - and it is dependent on acceleration. You can't determine what the actual direction of motion of an object is just by looking at the affect (e.g. something we can infer from the acceleration) , you have to look at the values which make a statement about the direction of motion (e.g. velocity, displacement etc.)

I'll try and avoid elaborating further than that, I'll be going away from the point I'm trying to make here.

[Bhootnike]

Okay, I'm not sure what you're asking here. Are you asking if gravity acts downwards?

hehe no :p
what i was asking was,  even though we assign one direction to be positive, and the other direction to be negative (e.g, let up = positive, let down = negative), do you ALWAYS just assume the force of gravity to be negative?
cause if you looked at those questions, and the working, if you assume that up is positive, down is negative, and that you always just assume gravity is negative, everything works out

just thought of my own lil example to further explain what exactly i mean

A person throws a ball up from a 50m high balcony, vertically upwards, at initial velocity of 10m/s, 20m from balcony vertically.
find the velocity with which the cricket ball hits the ground.

so for this:
Let UP = positive, DOWN = negative
easiest way is to look at the downwards journey from balcony.
so, our a= which is directed downwards, = -ve

u= -20
v=?
t=?
x(displacement)= -50m
a= -10

v^2 = u^2 + 2ax
       = 400 + 1000
       =1400
 v    =37.42 m/s


b) find time taken to reach the max height of 20m:
watch what i do with my variables:

u = +20m/s (since it is going upwards)
a = +10m/s^2 (since it is going upwards, and we assigned up as being positive, thus a  is positive)
x= +20m (as above)
t= ?
v=0

v= u  + at
t= (v-u)/a
t= -20/10
 = -2

which is obviously not right..

however, if we just ignored the whole, up = positive, down = negative bit JUST FOR ACCELERATION, and pretty much 'assumed' a due to g is ALWAYS negative,
then:
t= -20/-2
t = 10 secs

----------------------------------------------------------

Now, when an object is going upwards against gravity, we would usually say thats negative. but since i've explicitly defined up as being positive, then that wouldn't hold would it?


--------------

i've tried to grasp your second bit, but it just ain't coming to me. im missing something big hahah.. - ive either read too much and just made this too complicated or yeash, ive forgotten something.
and dw, i know your right, because otherwise the answers wouldnt comply with the solutions given to me! but i just need to get this !

pretty much with that question i gave in my other post, what i meant was:
in a) the ball is going UP, and if you assign up = positive, as theyve done in the question, then a should be positive and all other vectors should be +ve.
in d) the ball is going DOWn, and assigning down = negative, as theyve done in the question, then a should be negative, and all other vectors should be -ve.

[nerdgasm]

This is all paraphrasing what was said earlier, but for what it's worth, I'll attempt an explanation here.

what i was asking was,  even though we assign one direction to be positive, and the other direction to be negative (e.g, let up = positive, let down = negative), do you ALWAYS just assume the force of gravity to be negative?

Whether gravity is positive or negative depends on how we have assigned our directions.

Let's think about the common saying, "What goes up, must come down".
Gravity causes stuff to come down. If there were no gravity, we'd all be floating in the air! So, the force of gravity acts downwards.

So, gravity is positive or negative depending on which direction we've assigned to "up" and "down". If "up" is positive and "down" is negative, gravity is negative. If "up" is negative, and down is positive, then gravity is positive.

cause if you looked at those questions, and the working, if you assume that up is positive, down is negative, and that you always just assume gravity is negative, everything works out

Yes, that's true, because the questions assume up is positive, and down is negative. This happens because it's just the most natural way of looking at things. For example, the bottom floor of a building is the  ground (0) floor, then we go up to the 1st, 2nd, 3rd floors, and so on. Or think about a graph - the higher we go on the graph, the more positive the values get. The lower we go on the graph, the more negative the values we get.

BUT, (and this is a key point I will be stressing later), if I move from -10 to -5, I've moved upwards. If I move from 300 to 200, I've moved downwards.

Now, when an object is going upwards against gravity, we would usually say thats negative. but since i've explicitly defined up as being positive, then that wouldn't hold would it?

As others have tried to point out, acceleration and velocity DO NOT have to be in the same direction. Imagine I've just thrown a ball, at +20m/s into the air. Also, up is positive and down is negative, this does not change for the rest of the example.
The ball will get slower and slower until it reaches its maximum height (when it has no velocity at all). So, the ball's velocity has gone from +20m/s to 0m/s.

Using v = u + at, 0 = 20 + at. Clearly, time is positive. But that means that "a" has to be negative, or else the equation can never work. So, our velocity has been positive, but our acceleration has been negative.

The ball now starts to fall. It gets faster and faster until it reaches me again. Its velocity has gone from 0m/s to -20m/s (the negative indicates it is travelling downwards). Again, we use v = u + at. So, -20 = 0 + at, so -20 = at. But again, t must be positive, so a must be negative.

So, the ball was accelerating downwards, both when it was going up, and when it was going down.

As an alternative explanation, recall that  acceleration = change in velocity/change in time.
We always assume change in time is positive. Therefore, if change in velocity is positive, acceleration is positive. If change in velocity is negative, then acceleration is negative.

Draw a velocity-time graph for the situation described above. The graph starts at +20, then decreases at a constant rate, to 0, then decreases at a constant rate to -20. So, how has the velocity changed? Has it gone up or down? If it has gone up, then acceleration is positive. If it has gone down, then acceleration is negative. The actual numbers do not matter here. If you start at 1000 m/s, and end at 990 m/s, then acceleration is negative. If you start at -10m/s, and end at -5m/s, then acceleration is positive. What matters is the change in velocity.

I also encourage you to think in terms of forces. To give a really general example, say you're walking around at a constant speed, and I suddenly push you to the left. It doesn't matter whether you were walking left, right, backwards, forwards, (even up or down!), or any combination of those. Your acceleration would still have been to the left, because that's the direction of the force I applied.

Now, think about gravity. You're in the air. It doesn't matter whether you're going up or down. Your acceleration is still downwards, because that's the direction of the force gravity applies.

[Bhootnike]

Nice explanation. 
whilst reading ive grasped some things, but then, i tried another question and it hit me again haha...
take a look, maybe i can prove my madness now :p - EDIT: please try and point out what im thinking wrong here
-----------------------------------------------------------
- you can skip straight to c) if you want, i just included a and b to make the context clearer and see how the person has defined everything.

a projectile is projected at an initial velocity of 100m/s, at angle of 30degrees, from the top of a 120m tower:
calculate:
a) time taken to reach max height
b)max height above ground reached
c) time of flight

OK so, this bit isnt a part of official answer, but this is my understanding: up is positive, down is negative

u = 100sin30  = 50m/s        ----> this is going to be positive
a = -10m/s^2           - negative since g is down, and we've assigned down as negative
v=0
t= ?
x= ?

so : 0 = 50 + -10t
t = 5 secs

b) max height reached:
x=?
a=-10
v=0
u=50
t=5
0 = 2500 + -20x
x= 125

I'm good so far, its the next bit that troubled me:

c) total time: we know as per symmetry, time from top of tower, to max height = 5seconds. thus time for the projectile to reach a displacement of 0 again is just 10seconds.
so now we have to find the time it takes from the top of tower to the ground.
in the solutions its: (looking at vertical components)
 u = +50
a = -10
t=?
x= -120m

so doing math, and ignoring negative value obtained using null factor law, we get t = 12 secs

My trouble here is that, we assigned UP as positive, DOWN as negative.
so it makes sense that a = -10
also makes sense that x = -120m (i.e. displacement from the top of the tower, downwards to the ground, thus in downwards direction = NEGATIVE )
but u would be negative wouldnt it ... ?   i dont get why he's made u positive... , because we've assigned up as being positive.

is my problem here: am i meant to be looking at 'direction' with velocity? i.e. what my thought process is, is that: downwards journey is in the downwards direction, so therefore, velocity is negative.

just fyi: you said :

As others have tried to point out, acceleration and velocity DO NOT have to be in the same direction.

just letting you know, my decision to say velocity is in downwards direction is not influenced by the acceleration in this case!:)

[nerdgasm]

First of all, you are definitely not going crazy! Your physics has been good and correct!

What has happened, is that for part c), you have used an alternative method to the worked solutions!

The worked solutions are analysing ALL of the projectile's motion (from the very start, when it was launched). Basically, they're saying, 'Horizontal velocity doesn't affect our calculations to part c). Therefore, this is exactly the same as just throwing a ball off a 120m roof, with initial velocity of 50m/s upwards!' Then they use the constant-acceleration formulae to get an answer of 12 seconds. Their velocity is positive, because when you launch the projectile, it's going upwards.

Let's use your method. You have already worked out that the upwards half of the journey is 5s (in part a), and have used symmetry to work out that when the projectile is at roof height again, the time must be 10s.

We now have, v = -50m/s (by symmetry)
                       a = -10m/s^2
                       x = -120m (excellently reasoned)
                       t = ?

Again, using the constant acceleration formula x = ut + (at^2)/2, we obtain the positive solution t = 2. So, the projectile falls for a further 2 seconds, below the level of the roof. But it's already been travelling for 10s (as you've correctly worked out), which also comes to 12 seconds!

As you can see, both your method and the worked solutions yield the same answer, you've just gone a different pathway.

[Bhootnike]

First of all, you are definitely not going crazy! Your physics has been good and correct!

What has happened, is that for part c), you have used an alternative method to the worked solutions!

The worked solutions are analysing ALL of the projectile's motion (from the very start, when it was launched). Basically, they're saying, 'Horizontal velocity doesn't affect our calculations to part c). Therefore, this is exactly the same as just throwing a ball off a 120m roof, with initial velocity of 50m/s upwards!' Then they use the constant-acceleration formulae to get an answer of 12 seconds. Their velocity is positive, because when you launch the projectile, it's going upwards.

Let's use your method. You have already worked out that the upwards half of the journey is 5s (in part a), and have used symmetry to work out that when the projectile is at roof height again, the time must be 10s.

We now have, v = -50m/s (by symmetry)
                       a = -10m/s^2
                       x = -120m (excellently reasoned)
                       t = ?

Again, using the constant acceleration formula x = ut + (at^2)/2, we obtain the positive solution t = 2. So, the projectile falls for a further 2 seconds, below the level of the roof. But it's already been travelling for 10s (as you've correctly worked out), which also comes to 12 seconds!

As you can see, both your method and the worked solutions yield the same answer, you've just gone a different pathway.

thats comforting then

o ok haha.. TRUE! Time for vertical = time for horizontal. , and the reason they couldnt just do v=x/t, was cause they don't know the horizontal distance right?
thus, they worked with vertical. ahh makes sense!


okay makes sense thanks dude/gal !
appreciate it!
reason im so pedantic is cos i wanna show correct physics in my working, so yer
thanks!