Understanding Products in the Equation v^2 = u^2 + 2aS

[johncena]

In the equation v^2 = u^2 + 2aS , What kind of products are v^2 , u^2 , and aS ?
Cross product or dot product?

 

[Ask1122]

Please correct me if i am wrong, but i believe they will be dot products, cross products will have a change in directions as well.

 

[Fightfish]

Definitely dot products.
When you cross a vector with itself, you get the zero vector, which is absolutely meaningless.

 

[johncena]

OK . So v^2 and u^2 are dot products ...but what about aS?

 

[Fightfish]

OK . So v^2 and u^2 are dot products ...but what about aS?

Going from just a shallow point of view (without analyzing the meaning of the equation whatsoever) it must be a dot product as well as v^2 and u^2 are both scalars, which necessarily requires the product aS to yield a scalar as well.

 

[D H]

Unless of course a and S are already scalars.

 

[Gerenuk]

I think it is important to not just guess what the products might be, but rather prove the law anew. It might be none of the products. So let's do that
$$\Delta E_\text{kin}=\int\vec{F}\cdot\mathrm{d}\vec{s}$$
$$\therefore m|v|^2-m|u|^2=2\vec{F}\cdot\Delta\vec{s}$$
if the force is a constant
$$\therefore |v|^2=|u|^2+2\vec{a}\cdot\Delta\vec{s}$$
or if you wish
$$\therefore \vec{v}\cdot\vec{v}=\vec{v}_0\cdot\vec{v}_0+2\vec{a}\cdot(\vec{s}-\vec{s}_0)$$

Note that all this assumes that the force/acceleration is constant.

 

[Redbelly98]

Note that all this assumes that the force/acceleration is constant.

[nitpick]

Just to clarify, Gerenuk means that force and acceleration are both constant.

[tex]\frac{force}{acceleration}[/itex] is the same as the mass, which is always constant (at nonrelativistic speeds)<br /> <br /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> [/nitpick][/tex]