Physics review: Vector quantities

OK... I decided to write up something for any potential GAMSAT sitters reading this site.

Quite commonly when someone begins learning physics, and doing problems, they sub numbers into formulas to get the answer. Yes, that's what you're supposed to do for numerical answers to get the solution, although sometimes there may be a few steps involved rather than just one step.

However, a common elementary mistake is to forget about directions. When we are dealing with vector quantities, we must keep in mind not only the magnitude, but also the direction. Otherwise, your answers will come out weird. Just as an illustration, if I'm facing south and initially travelling south at 3 m/s then turn around and travelling north at 6 m/s, it's different from if I was initially travelling north at 3 m/s and just sped up to 6 m/s. In the first instance, my change in velocity was 9 m/s north (i.e. 6-(-3), or 6+3), but in the second instance it was only 3 m/s north (i.e. 6-3).

In two or three dimensions, there are a few ways to approach vector problems. You can draw vectors and add them "head to tail", etc, and/or separate into orthogonal components and deal with each component separately. For one dimension, the main thing required is just to assign one direction as positive and label the signs of all variables accordingly (ie, opposite direction means that the variable's value is negative).

I'll just finish with an example now. Perhaps I'll just use the same example as I started off with.

"A person was travelling south at 3 m/s initially. He slows down, then turns around and then runs north, reaching 6 m/s. He does all this in 3 seconds. What is his average acceleration?"

Well, in this case, I'll assign north to be positive.

His initial velocity is south 3 m/s, and since north is positive, south is the opposite direction and is negative. So u=-3 m/s. (u=initial velocity)

He is finally travelling north, which was assigned as positive. So v=+6 m/s.

"Time" is 3 seconds; t=3 s (time is actually a scalar quantity, so usually it should be positive)

Anyway, so a=(v-u)/t=(+6-(-3))/3
=9/3
=3 m/s²

Since this is positive, and we assigned north as positive, this means that the acceleration's direction is north. Technically the acceleration needs both a magnitude and a direction; so just saying it's 3 m/s² isn't enough. Anyway, so from that, acceleration is 3 m/s² north.


Now, actually it still works if we assigned south as positive; you don't have to worry about which direction to assign as positive in two directions too much, although it's good to choose one where you don't have too many negatives, since it makes things easier.

Anyway, if I assigned south as positive instead (just to be different),

initial velocity=u=3 m/s south=+3 m/s (since south is positive)
final velocity=v=6 m/s north=-6 m/s (since north is opposite to south, where south is positive)

time=3 s

a=(v-u)/t
=(-6-3)/3
=-3 m/s²

However, since south is positive, and the answer is negative, that means acceleration is 3 m/s² north (ie, since south is positive, north is negative in this instance; and we have a negative answer). Same as before, 3 m/s² north, as long as you interpret the sign correctly by the constraints you gave initially.

A common mistake might be to just use magnitudes; forgetting that directions are important. Doing this:
u=3 m/s
v=6 m/s
t=3 s

a=(v-u)/t=(6-3)/3=1 m/s²; different from the correct answer of 3 m/s² north!

Plugging in magnitudes without thinking about directions only gives the correct answer usually when the direction is unchanged; ie if the person was initially going 3 m/s north and accelerated to 6 m/s north, or initially 3 m/s south and accelerated to 6 m/s south in 3 seconds, then the acceleration's magnitude would be 1 m/s². Not for the question given above though!

So... when doing physics and plugging numbers into formulas, remember to take into account directions! In 1D, this means assigning a positive direction, and figuring out whether a value takes a positive or negative sign, before solving the algebra for the answer. Otherwise, if you've got a few quantities in different directions, your answer is probably going to turn out wrong.