Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove that v dot a= uu'

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data
    the question is simply asking to prove that if speed is constant than velocity and acceleration vectors are perpendicular to each other. it also says as hint to differentiate v dot v= u^2.....
    V=velocity A=acceleration u=speed t=time

    2. Relevant equations
    i suppose knowing the vector dot product properties would be useful to have around,so heres some:
    V dot V= [v]^2(magnitude)
    derivative of vector dot products:
    d(a dot b)/du= da/du dot b+ a.db/du
    acceleration is= the derivative of Velocity with respect to time

    3. The attempt at a solution

    ok so ive used the hint and tried to differentiate V dot V= u^2 and i get

    dv/dt dot V+V dot dv/dt= du/dt u^2= 0 since speed is constant
    2V dot dv/dt=0
    now im not sure if this is representinvg acceleration or not the 2v dv/dt or if its the rate of change in speed or whatever, if its acceleration then wouldnt it be 2A=0, but now it doesnt make sense since acceleration cant be 0 so i know theres a mistake somewhere...

    my other idea was to use V dot V=[V]^2 which is one of the dot product properties
    and then the magnitude of the velocity would be the speed wouldnt it?

    dv/dt [V]^2=0

    anyways im lost,so can someone please help?
  2. jcsd
  3. Sep 17, 2012 #2


    User Avatar
    Homework Helper
    Gold Member

    acceleration (a vector!) is the time derivative (rate of change) of the velocity vector, so [itex]\mathbf{v} \cdot \frac{d \mathbf{v} }{ dt } = \mathbf{v} \cdot \mathbf{a} = 0[/itex]

    Under what conditions will two vectors have a dot product of zero?
  4. Sep 17, 2012 #3
    ok thats what i thought dv/dt= A but its not necessarily 0, if they( vector V and Vecotor A) are perpendicular to eachoter the dot product would be zero...

    ok so V dot A=0
    so when i first derived V dot V= u^2 and got to dv/dt V+V dv/dt= u^2 du/dt=0

    du/dt is the rate of change of speed correct? so if the first derivative of u is =0
    would't u'= equal zero?

    V dot A= uu' -------------> V dot A= u(0)?
  5. Sep 17, 2012 #4
    or is it better to make V dot V= [V]^2

    sqrt [V]^2= sqrt u^2 ---------> [v]= u

    d[v]/dt [v]= du/dt u = 0 --------> d[v]/dt = [a] magnitude of the acceleration
  6. Sep 17, 2012 #5


    User Avatar
    Homework Helper
    Gold Member

    It is better to use [itex]v^2= |\mathbf{v}|^2[/itex] than [itex]u^2[/itex], since [itex]\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2[/itex] follows from the definition of dot product, and by definition, the speed is the magnitude of the velocity.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook