Why is the force and acceleration positive?

[starstruck_]

Hi! So someone came up to me with this question during physics and its been bothering me because I didn't know what to say to them and I didn't know if my understanding is correct or not. The question was something like:

" A 72 kg car is traveling at v=63km an hour when it gets into a collision, coming to a stop in d= 14 m.

Calculate the magnitude of the net force applied on the car. When I did my whole equation rearranging stuff using 0= v1^2+2(-a)d

I got a= v1^2/(2a) and hence, force = ma= m(v1^2/(2d)).
When I did it I was like okay great got everything I need, I'm good to go so I didn't really think about the signs of my answers, so I really didn't knoe what to say to the person who asked me why the signs were positive when, when you're stopping and set the direction of your displacement as positive, your force and acceleration come put negative even though they're opposing your motion.

I tried to give an answer in that we've already accounted for the fact that acceleration is negative in our initial expression, but if we were to put it back into v2^2=v1^2+2ad we would put in the negative of the answer and that we changed that plus sign to a minus sign so we accounted for the negative direction of the acceleration and just need the positive magnitude? I don't know.

It made sense to me mathematically so :/

Can someone help out? This is really bothering me for some reason

 

[Chestermiller]

The acceleration is in the negative x direction, and the net force is also in the negative x direction. If you write the force as a vector involving the unit vector in the x direction, what do you get?

 

[robphy]

I think a clearer notation in terms of vectors and components might be helpful.

Let's assume 1-d motion in the x-direction.
There may be some confusion distinguishing the "magnitude of a vector" from the "x-component of the vector" when dealing with 1-d motion.
In [itex]v_f^2=v_i^2+2a\Delta x [/itex],
is the symbol "[itex]a[/itex]" the x-component [itex]a_x[/itex] or the magnitude [itex] \|\vec a\| [/itex]?

[itex]v_f^2=v_i^2+2a\Delta x [/itex],
is really a vector equation
[itex]v_f^2=v_i^2+2\vec a\cdot \Delta \vec x [/itex],
which can be written in terms of components
[itex]v_f^2=v_i^2+2 a_x \Delta x [/itex]For an object traveling in the positive-x direction (so [itex]\Delta x >0[/itex] and [itex]v_x>0[/itex]),
an acceleration [itex] a_x<0[/itex] decreases the velocity [itex]v_x[/itex] and
(because [itex] a_x v_x<0[/itex]) decreases the speed [itex]v=|\vec v|[/itex].
When you express [itex]a_x[/itex] in terms of the magnitude of acceleration vector, then [itex]a_x=-\| \vec a\| [/itex].

Similarly, instead of [itex] F_{net}=m a[/itex]
write
[itex]\vec F_{net}=m \vec a [/itex]
or
[itex] F_{net,x}=m a_x[/itex].

(Interpreting [itex] F_{net}=m a[/itex] as magnitudes
means [itex] \| \vec F_{net}\|=m \| \vec a\| [/itex], which is true, but says nothing about directions.)

 

[A.T.]

I really didn't knoe what to say to the person who asked me why the signs were positive

Positive signs in some equation don't imply positive numerical values. For example: X = Y doesn't imply that X and Y are positive, just that they have the same sign. They could both be negative as F and a are in your case.

 

[Chestermiller]

$$\vec{a}=\left(-\frac{v_i^2}{2d}\right)\vec{i_x}=\left(\frac{v_i^2}{2d}\right)(-\vec{i_x})$$
$$\vec{F}=m\vec{a}=m\left(-\frac{v_i^2}{2d}\right)\vec{i_x}=m\left(\frac{v_i^2}{2d}\right)(-\vec{i_x})$$

 

[starstruck_]

Positive signs in some equation don't imply positive numerical values. For example: X = Y doesn't imply that X and Y are positive, just that they have the same sign. They could both be negative as F and a are in your case.

Right so basically all I really got from solving the equation was the magnitude for my acceleration and force, because I already accounted for the fact that my acceleration was negative earlier?

 

[Chestermiller]

Right so basically all I really got from solving the equation was the magnitude for my acceleration and force, because I already accounted for the fact that my acceleration was negative earlier?

Did my post #5 not work for you? What was not clear?

 

[starstruck_]

Did my post #5 not work for you? What was not clear?

Sorry, I wasn't able to see the symbols off mobile, but now that I can see them, so you've taken acceleration and multiplied it with a unit vector in the negative x direction? Which is the same as using a unit vector in the positive x direction, making acceleration negative?

Does this mean, that once I plugged in "-a" into the equation, I had essentially multiplied acceleration with a unit vector in the negative x direction, hence resulting in a positive a value?

 

[Chestermiller]

Sorry, I wasn't able to see the symbols off mobile, but now that I can see them, so you've taken acceleration and multiplied it with a unit vector in the negative x direction? Which is the same as using a unit vector in the positive x direction, making acceleration negative?

Does this mean, that once I plugged in "-a" into the equation, I had essentially multiplied acceleration with a unit vector in the negative x direction, hence resulting in a positive a value?

Looking st the equations I wrote for the acceleration vector and the force vector, how could you possibly think that these vectors are pointing in the positive x direction? And , when you are talking about the magnitude of a vector, are you unaware that the magnitude is always regarded as positive?

 

[Mister T]

Right so basically all I really got from solving the equation was the magnitude for my acceleration and force, because I already accounted for the fact that my acceleration was negative earlier?

You're right. The correctly-written version of the equation you used is ##v_x^2=v_{xo}^2+2a_x\Delta x## because it's a relationship between the x-components of the velocity and acceleration vectors. Thus you substituted ##-a## for ##a_x## where ##a## is some positive constant, a scalar. It is really this mix up between vectors, vector components, and vector magnitudes that is the source of the confusion you experienced. Few textbook authors make this distinction clear, probably because the notation is cumbersome when you insist on putting a subscript of ##x## on all those variables. One introductory physics textbook author who is conscientious about this is Randy Knight. You can find his textbook on the web, despite the fact that it's a commercial publication.

 

[starstruck_]

You're right. The correctly-written version of the equation you used is ##v_x^2=v_{xo}^2+2a_x\Delta x## because it's a relationship between the x-components of the velocity and acceleration vectors. Thus you substituted ##-a## for ##a_x## where ##a## is some positive constant, a scalar. It is really this mix up between vectors, vector components, and vector magnitudes that is the source of the confusion you experienced. Few textbook authors make this distinction clear, probably because the notation is cumbersome when you insist on putting a subscript of ##x## on all those variables. One introductory physics textbook author who is conscientious about this is Randy Knight. You can find his textbook on the web, despite the fact that it's a commercial publication.

I just tried working with the equation without subbing in a negative acceleration right away and wow I'm dumb, this makes sense! Thank you! I should probably open up my linear algebra textbook some day, I know what unit vectors are but uh I don't remember anything to do with them.