# Suvat vector versus the scalar form

• heroslayer99
In summary: Well...I'll try to find some even higher authority... NIST, SI, ???Temperature, energy, work, and power can all be negative.
kuruman said:
Suppose one has ##\vec A = 3 \,\hat i - 4 \,\hat j##. I am using specific numbers instead of ##A_x## and ##A_y## to remove the sign ambiguity that is inherent in algebraic symbols.

I see two interpretations for the term ##- 4 \,\hat j## representing the vector component of ##\vec A## in the y-direction.
1. A vector of negative y-component pointing in +y-direction, i.e. ##- 4 \,\hat j=(-4)(+\hat y).##
2. A vector of positive magnitude 4 pointing in the negative y direction, i.e. ##- 4 \,\hat j=(+4)(-\hat y).##
We teach students to say that vector ##\vec A = 3 \,\hat i - 4 \,\hat j## has a negative y-component, which conforms with interpretation 1, yet in everyday life we conform with interpretation 2. When someone asks us, "which way did he go?", we extend our arm to form a unit vector with its tail at our shoulder and its tip at the end of our index finger and say, "he went that-a-way" whichever way that is. If he went South, we say and point "South" (interpretation 1), we don't say "North moving backwards" (interpretation 2.) Small wonder there is confusion.

Just a few thoughts ##\dots##
I’d argue that going 4 m in one direction would be equivalent to going -4 m in the opposite direction. That would eliminate any dependence on interpretation.

heroslayer99 said:
Ok and just because I am pedantic, suvat in 1D is with the components of the vectors in that direction (which is a scalar and hence +ve or -ve)
If you have an equation like ##\vec{v}=\vec{v_o}+\vec{a}t## then you can write ##v_x=v_{ox}+a_xt##. What we are saying is that if two vectors are equal then their x-components are also equal. We usually write the latter equation as ##v=v_o+at## for brevity, but it can be a source of confusion for many students.

Orodruin said:
I’d argue that going 4 m in one direction would be equivalent to going -4 m in the opposite direction. That would eliminate any dependence on interpretation.
You argue correctly, of course. However, the equivalence is not at all reinforced by day to day usage in everyday life or in statements of physics problems. Even though there is equivalence, we say, "the pirate treasure is buried 4 m North of this tree"; we don't say, "the pirate treasure is buried - 4 m South of this tree." Likewise, in a physics problem, when we want to provide magnitude and direction, we say "the velocity of car A is 4 m/s towards car B"; we don't say "the velocity of car A is -4 m/s away from car B." Despite the equivalence, humans have an aversion for negative signs which introduces an asymmetry in usage to avoid the confusion that might arise from a double negative.

My point is that this asymmetry leads one to think of a vector component of a vector, i.e. ##-A_y~\hat y## as a magnitude ##|A_y|## times a direction ##(-\hat y)##. Up to this point that's OK. But then comes the physics teacher who says that ##A_y##, the y-component of vector ##\vec A##, is a scalar. At this point one erroneously concludes "therefore scalars are "always" positive."

When vectors are first introduced, I think that usage of the term "scalar" should be avoided even if accompanied by a relaxed definition, as @PeroK suggests in #29, because it invites trouble. A vector is usually introduced as a mathematical entity that has magnitude and direction. If then we define a scalar as an entity that has magnitude only but no direction, ##~\dots~## oops, I mentioned the M-word which is used for a positive quantity only.

In my opinion, there is no good way to introduce scalars properly at the intro level. If I stop and think about it, I see no harm done by not mentioning scalars at that level either.

PeroK
At the same time, I still remember precisely going in the opposite direction to a given direction being how I was first introduced to negative numbers. I still have a vague memory of the illustration from the low level maths textbook somehow ingrained on my retina.

kuruman
carzzpeter Referenc said:
Which one of them? Each component of the suvat equation is a scalar equation.

If you "reply" to the post, we can see what you are referring to here. Otherwise, it's not clear what you're asking about.

PeroK said:

If you "reply" to the post, we can see what you are referring to here. Otherwise, it's not clear what you're asking about.
I think it is fairly safe to assume that replies of the form in #40 are directed directly at the OP without checking the rest of the thread to see what has already been said. Clarity would of course be best though.

heroslayer99 said:
I've been told earlier that a vector component is a scalar

vela said:
while Ax is the scalar component of A→ in the x-direction

heroslayer99 said:
displacement vector has a component of -3 in the i direction?

kuruman said:
Ay, the y-component of vector A→, is a scalar
As I understand it, a component of a vector is a vector. In the vector ##x\hat i+y\hat j##, ##x\hat i## is the ##\hat i## component, while the scalar x is the coefficient of ##\hat i##.

haruspex said:
As I understand it, a component of a vector is a vector. In the vector ##x\hat i+y\hat j##, ##x\hat i## is the ##\hat i## component, while the scalar x is the coefficient of ##\hat i##.
This may be a question of nomenclature. The use of ”component” for ##x## is widespread and i would consider it pretty standard. Just a random example from a quick internet search : https://www.grc.nasa.gov/www/k-12/airplane/vectpart.html

nasu
Orodruin said:
This may be a question of nomenclature. The use of ”component” for ##x## is widespread and i would consider it pretty standard. Just a random example from a quick internet search : https://www.grc.nasa.gov/www/k-12/airplane/vectpart.html
I regard that as unfortunate and confusing.

jbriggs444
haruspex said:
I regard that as unfortunate and confusing.
That may be so and it is your prerogative to consider it as such. However, it does not change the pretty standard usage.

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