Acceleration always perpendicular to velocity

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randomafk
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Homework Statement



If the acceleration/force vector is always perpendicular to the velocity vector, what is the path?

Homework Equations



F=ma
a dot v=0


The Attempt at a Solution



We know that the dot product of a and v is zero such that

vx*vx'+vy*vy'=0 where vx'=dvx/dt

Also, I know this would be UCM, and given that I could say that the speed is constant. However, how would you know speed is constant not knowing the final path?

In addition, extending this further, how would know what path a particle took given acceleration and velocity vectors. That is to say, what if acceleration and velocity were not always perpendicular and also changed in magnitude with time?
 
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tiny-tim said:
hi randomafk! welcome to pf! :smile:


v.v' = 0 :wink:


thanks for the welcome and help

if s=speed, then
s=sqrt(v.v)
How does this thing relate to v.v' = 0

Even if you knew that speed is constant, how could that demonstrate UCM?
 


randomafk said:
thanks for the welcome and help

if s=speed, then
s=sqrt(v.v)
How does this thing relate to v.v' = 0

Even if you knew that speed is constant, how could that demonstrate UCM?

What happens to the velocity vector ?
 
The velocity vector should stay constant in magnitude but change direction

tiny-tim said:
integrate :wink:

okay. so if i do an integral over dv

dv=a dt

[tex]\int[/tex]v.dv = [tex]\int[/tex] v.adt=C since v.a=0 for all t

I still can't figure how to link back to v.v=0