How do you calculate vector subtraction in physics?

[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 $$\arctan\frac{C_y}{C_x}$$.