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

AI Thread Summary
In the equation v^2 = u^2 + 2aS, v^2 and u^2 are confirmed as dot products since they are scalars. The term aS is also considered a dot product to maintain consistency in scalar output, assuming a and S are scalars. The discussion emphasizes the importance of proving the relationships rather than making assumptions about the products. It is noted that these calculations assume constant force and acceleration. The overall consensus is that all terms in the equation yield scalars, reinforcing the dot product nature of the products involved.
johncena
Messages
131
Reaction score
1
In the equation v^2 = u^2 + 2aS , What kind of products are v^2 , u^2 , and aS ?
Cross product or dot product?
 
Physics news on Phys.org
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.
 
Definitely dot products.
When you cross a vector with itself, you get the zero vector, which is absolutely meaningless.
 
OK . So v^2 and u^2 are dot products ...but what about aS?
 
johncena said:
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.
 
Unless of course a and S are already scalars.
 
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.
 
Gerenuk said:
Note that all this assumes that the force/acceleration is constant.
[nitpick]

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

\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]
 
Back
Top