Speed with constant acceleration

[ahuebel]

Using a Newton equation for position and substituting (v-v0)/a for t you can get the following formula:
v^2= v0^2+2a(x-x0) where v^2 is supposedly speed. I understood speed to be tha magnitude of velocity which is the sqrt of the sum of the squares of the components (i.e. sqrt(vx^2+vy^2+vz^2)). I seem to be missing something... how is speed then v^2? I am sure it is right in front of my face but I am just not seeing it.

TIA

 

[Doc Al]

v^2 is the speed squared; v is the speed.

Realize that your formula ,v^2= v0^2+2a(x-x0), is for uniformly accelerated straight-line motion.

 

[ahuebel]

Thank you. It just seems odd that if they would state that an equation is speed as a fucnction of position it would be much better to have that equation be v = sqrt(v0^2+2a(x-x0)). It can throw simple people like me off :).

 

[Doc Al]

Think of that equation as a relationship among all the different variables, not just as a formula for finding speed. Depending upon what information is given, you may wish to use it to solve for any of the variables. (And having a single v^2 is much nicer looking than that ugly square root! )