How do you calculate vector subtraction in physics?

[Byrgg]

Oh, wait a second, in the +5 case with zero being the final, -5 would be the change, right? And what do you mean by the total change in a two dimensional case? If you have some starting vector, and then some ending vector, is the resultant the change between those two? And in the one dimensional case(sorry for jumping around like this), I was thinking that, if you add +5 and -5, you get obviously get 0. What significance does this zero have, what is it, exactly? The final velocity?

If you have:

-->

and then you add:

<--

Then you get something like this:

-->
<--

When you add those two vectors, you get 0. I'm wondering what exactly this zero is. Sorry if this is unclear or anything. I'm sure you've already explained this in some earlier post, some way or another, but it just keeps on confusing me.

 

[Byrgg]

I"ll try to explain this a little better. Say you're traveling at 5 m/s [N], and then suddenly you start traveling at 5 m/s . Through vector analysis, wouldn't you have to change your velocity by 10 m/s , in order to achieve this? But on the other hand, can't you just turn around, and start moving in the opposite direction? How is this a change of 10 m/s ? Also, in order to achieve a velocity of 0 m/s, you would need to change your velocity by 5 m/s , right? Don't you just stop? How is stopping the same as changing your velocity 5 m/s ?

And, in my previous post, what exactly is 0? Is that the final velocity, whereas the negative vector, is just the change in velocity?

 

[Tomsk]

I think you're confusing two situations. The first one you gave was:

Initial velocity = +5m/s
Final velocity = -5m/s
Change in velocity = -10m/s

Remember,
Change = Final - Initial;
-10 = (-5) - (+5)

Then, just now you are adding +5 and -5 and getting zero, this is the situation:

Initial velocity = +5m/s
Final velocity = 0m/s
Change in velocity = -5m/s

Change = Final - Initial;
-5 = (0) - (+5)

You are not considering acceleration in either of these situations, sorry if I confused you about that.

 

[Byrgg]

So then, for what purpose are vectors ever added, if you subtract to determine change?

 

[Tomsk]

You could add them if you know the initial velocity and the velocity added, and you want the final velocity. It depends on what you want to find out and what you know.

 

[Byrgg]

You said earlier that acceleration isn't considered in either of these situations, I thought it would be, considering the fact that velocity is changing.

 

[andrevdh]

Correct, the object is accelerating if the velocity of the object is changing, but what he meant is that it is not calculated in these examples. All that is under discussion is the change in the velocity iat this stage (it is a difficult enough subject on its own). One would need to divide by the time interval between the two cases tot get the acceleration.

 

[Byrgg]

Ok, so let me try to understand this, if you're adding vectors, you're getting the final velocity? For some reason, I'm having a hard time understanding that adding 5 m/s [N] and 5m/s results in a final velocity of 0, the 5 m/s is the change, right? Does this still apply in two dimensional cases? I'm thinking that the final velocity may not be represented by the resultant vector of adding all of the vectors in a two dimensional case, but I'm probably wrong. Should I be thinking of vetors as a change in velocity, rather than an actual velcoity?

Also, in an earlier example, the equation C = A - B was supposed to represent the change in velocity, if B is the final vector, should the change be represented by C = B - A?

 

[andrevdh]

As an example consider someone trying to row upstream to get away from a waterfall, his final velocity unfortunately being zero, which means he will survive as long as he keep on rowing! - Sorry I'm off again to go and do some work.

 

[Byrgg]

Which of my questions was that directed at?

 

[andrevdh]

Ok, so let me try to understand this, if you're adding vectors, you're getting the final velocity? For some reason, I'm having a hard time understanding that adding 5 m/s [N] and 5m/s results in a final velocity of 0, the 5 m/s is the change, right? Does this still apply in two dimensional cases? I'm thinking that the final velocity may not be represented by the resultant vector of adding all of the vectors in a two dimensional case, but I'm probably wrong. Should I be thinking of vetors as a change in velocity, rather than an actual velcoity? ...QUOTE]

The flow of the water say being 5 m/s N and the rower rowing 5 m/s S. His resultant velocity will then be 0 m/s (relative to the embankment). That is he will be standing still, but if we consider another example of an object moving 5 m/s S and a little while later 5 m/s N, then the change in his velocity will be 10 m/s N in order to change from 5 m/s S to 5 m/s N.

 

[Byrgg]

Ok, I'm pretty sure I think I understand this now, but what about my other question?

Also, in an earlier example, the equation C = A - B was supposed to represent the change in velocity, if B is the final vector, should the change be represented by C = B - A?

 

[coolblue]

i got 12.6 m/s north of east. but i could be wrong

 

[jfsbird]

The first step when doing vector addition or subtraction is to find the component form of each vector. Components are related to x and y coordinates and are found using A_x=\vecA cos\Theta and A_y=\vecA sin\Theta . Once in component form you add A_x+B_x and A_y+B_y, this gives you components for the resultant vector. The magnitude of the resultant vector C is \sqrt{C_x^2+C_y^2}. The direction of the resultant Vector C is \arctan\frac{C_y}{C_x}.