• Support PF! Buy your school textbooks, materials and every day products Here!

Prove that v dot a= uu'

  • #1

Homework Statement


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


Homework 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


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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6

Homework Statement


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


Homework 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


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?
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?
 
  • #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)?
 
  • #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
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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.
 

Related Threads on Prove that v dot a= uu'

Replies
19
Views
7K
Replies
3
Views
6K
  • Last Post
Replies
0
Views
925
  • Last Post
Replies
9
Views
15K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
2
Views
997
Replies
3
Views
5K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
5K
  • Last Post
Replies
2
Views
413
Top