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

Velocity of a particle GIVEN position vector!

  • Thread starter vorcil
  • Start date
391
0
A [article follows the path given by the position vector
r= (4t , 3 , t^3)

what's its' speed at t=1???

no idea how to solve this
i know velocity is the derivative of the position with respect to time

do i just solve, 4, 0, 3t^2, then stick it in?
4 + 3 = 7m/s??
 
Velocity is a vector quantity. The particle's velocity at time t is v = (4,0,3t^2). At time t=1, the velocity of the particle is v = (4,0,3). It's speed at time t=1 is [tex]\sqrt{4^{2}+0^{2}+3^{2}}[/tex] = 5.
 
391
0
Velocity is a vector. The particles vecolocty at time t is v = (4,0,3t^2). At time t=1, the veclocity of the particle is v = (4,0,3). It's speed at time t=1 is [tex]\sqrt{4^{2}+0^{2}+3^{2}}[/tex] = 5.

oh so it's just the magnitude of the differentiated vector?

this was the original equation
r= (4t , 3 , t^3)

so i'd differnetiate it,

(4 , 0 , 3t^2)
and find the |r| or the magnitude of it?
 
EDIT: You want the magnitude of v, not r.
 
391
0
Yes. Speed is a scalar.
I find it strange to differentiate the position vector

what happened to the square on the t^3, 3t^2, it became 4,0,3?
what about the t^2? or is the ^2 part of the t, and we forget the t's?
 
You wanted the speed at time t=1.
 
You're given the particle's postion as a function of time. If you differentiate, then you know the rate and direction at which the particle's position is changing.
 
391
0
You wanted the speed at time t=1.
so how do i make r= (4t , 3 , t^3) become 4,0,3?
4*1t , 3*0t, 1*1^3?

i thought differentiating r= (4t , 3 , t^3) 4t -> 4 ,3 -> 0, t^3 -> 3t^2
then substituting in 1 for t, you get the velocity vector, 4,0,3?
then i'd just get |(4,0,3)|
 
32,838
4,562
v = dr/dt = (4, 0, 3t2)

speed = |v| = [itex]\sqrt{4^2 + 0^2 + 9t^4 }[/itex]
The speed at t = 1 is the value of the radical above at t=1.
 
so how do i make r= (4t , 3 , t^3) become 4,0,3?
4*1t , 3*0t, 1*1^3?

i thought differentiating r= (4t , 3 , t^3) 4t -> 4 ,3 -> 0, t^3 -> 3t^2
then substituting in 1 for t, you get the velocity vector, 4,0,3?
then i'd just get |(4,0,3)|
(4,0,3) is a vector with magnitude 5.
 
391
0
Velocity is a vector quantity. The particle's velocity at time t is v = (4,0,3t^2). At time t=1, the velocity of the particle is v = (4,0,3). It's speed at time t=1 is [tex]\sqrt{4^{2}+0^{2}+3^{2}}[/tex] = 5.

OH RIGHT,
thanks Random variable, XD XD XD :rofl:

Just to confirm, i differentiate r= (4t , 3 , t^3)?????
 

Related Threads for: Velocity of a particle GIVEN position vector!

Replies
7
Views
2K
Replies
11
Views
5K
Replies
4
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
2
Views
689
Replies
4
Views
1K
Replies
9
Views
6K

Hot Threads

Top