Do these equations belong to kinematics or dynamics? Or both, maybe?

[BruceW]

(you can write it either way, they both mean the same thing, but physicists often write it the way in your textbook, because its easier)
lazy so-and-so's...

 

[Femme_physics]

it means:
$$( \cos( \alpha ) )^2$$
The notation is difficult to get used to at first

Thanks :) (missed that reply)

Another question with respect to these formulas.

Sometimes I see "g" with a minus before it. Sometimes I see "g" without a minus before it. If I decide to plug "g" as a minus-- where there is already a minus it's supposed to cancel-out, and where there is no minus it's actually plugged as a minus-- according to the formula. Or, do I always plug "g" without the sign, and then it accepts whatever sign the formula gives it?

 

[I like Serena]

Sometimes I see "g" with a minus before it. Sometimes I see "g" without a minus before it. If I decide to plug "g" as a minus-- where there is already a minus it's supposed to cancel-out, and where there is no minus it's actually plugged as a minus-- according to the formula. Or, do I always plug "g" without the sign, and then it accepts whatever sign the formula gives it?

Always accept the sign the formula gives for "g".

It should help if you start understanding the meaning of "g" and in particular in which direction it "works".

 

[Femme_physics]

Thanks :) Great, will do!

 

[rock.freak667]

Thanks :) (missed that reply)

Another question with respect to these formulas.

Sometimes I see "g" with a minus before it. Sometimes I see "g" without a minus before it. If I decide to plug "g" as a minus-- where there is already a minus it's supposed to cancel-out, and where there is no minus it's actually plugged as a minus-- according to the formula. Or, do I always plug "g" without the sign, and then it accepts whatever sign the formula gives it?

It depends on the convention used. For example if you have an object falling under gravity, so you only need to consider up and down motion, so the particle is moving down, you can take down as positive and use g = + 9.81 m/s².

If you take up as positive then g acts downwards and thus g = -9.81 m/s².

In cases where the formula is like v=u-gt, then down is negative and the minus sign covers for that, so you'd just put in g= 9.81 into the equation.

 

[I like Serena]

It depends on the convention used. For example if you have an object falling under gravity, so you only need to consider up and down motion, so the particle is moving down, you can take down as positive and use g = + 9.81 m/s².

If you take up as positive then g acts downwards and thus g = -9.81 m/s².

In cases where the formula is like v=u-gt, then down is negative and the minus sign covers for that, so you'd just put in g= 9.81 into the equation.

I'm sorry, I'm going to have to disagree here.

"g" is a physical constant and never has a sign. It is just g = 9.81 m/s².

This is opposed to usage of for instance a generic acceleration "a" of which the direction is unspecified, which might for instance be -g or +g.

If you look at the formulas in the first post, you'll see any sign of "g" is definitely accounted for! ;)

 

[Femme_physics]

I thought rock.freak's reply had somewhat contradicted yours, but yours did make more sense to me. I appreciate anyone's feedback, regardless! :)

 

[rock.freak667]

I'm sorry, I'm going to have to disagree here.

"g" is a physical constant and never has a sign. It is just g = 9.81 m/s².

This is opposed to usage of for instance a generic acceleration "a" of which the direction is unspecified, which might for instance be -g or +g.

If you look at the formulas in the first post, you'll see any sign of "g" is definitely accounted for! ;)

My post was supposed to make the point that your sign for 'g' would depend on what direction you use as positive, not that 'g' will be -9.81 m/s² regardless of direction chosen.

 

[Doc Al]

My post was supposed to make the point that your sign for 'g' would depend on what direction you use as positive, not that 'g' will be -9.81 m/s² regardless of direction chosen.

The point that I Like Serena was making is that while the acceleration can be +g or -g depending on your sign convention, g itself is just the magnitude of the acceleration due to gravity. It's always positive (unsigned, actually).

 

[BruceW]

"g" is a physical constant and never has a sign. It is just g = 9.81 m/s2

This is true, but if she gets an exam that explicitly gives g = -9.81 m/$s^2$, and then this equation (for example):
$$depth = \frac{gt^2}{2}$$
Then she should use the given value for g.
In other words, the notation which the exam gives is the most important.
So here, the examiner is looking for a negative value for depth, but if she used g as positive, then she'd get an incorrect positive value for depth. (Incorrect in the sense that it doesn't conform to the examiner's notation).

 

[Femme_physics]

Eep! I hope the system would do its best to not confuse me!

Thanks for the heads up, and the input! :)

 

[nrqed]

Eep! I hope the system would do its best to not confuse me!

Thanks for the heads up, and the input! :)

Bonjour!

J'espère que la préparation pour tes examens se passe bien!



Some people do use g to be -9.81 m/s^2 although they are the exceptions.
But I can tell from your formula sheet that your class is using g = 9.81 m/s^2.



So if your teacher and textbook are consistent, you should be using a positive value all the time, with no exception.

Now, if sometimes your teacher or your book chooses to put the positive y-axis pointing down, all the signs of the velocities (V, Vi) as well as the acceleration will flip signs in the equations, but g will remain positive. Hopefully your teacher does not do that as it is always confusing to students. I avoid doing this in my classes.

Bonne chance dans tes études!