Deriving Velocity and Acceleration from Position Vector r(t)

  • Thread starter Thread starter GoodTimes
  • Start date Start date
  • Tags Tags
    Derivation Vector
AI Thread Summary
The discussion focuses on deriving velocity and acceleration from a given position vector r(t) for a particle moving along a space curve. The user initially attempts to differentiate the position vector to find the velocity and acceleration but is unsure about the correctness of their calculations. Responses clarify that the differentiation approach is correct, but one derivative contains an error that needs to be corrected. The use of LaTeX for mathematical notation is also suggested for clarity. Ultimately, the user confirms their task is to find the velocity and acceleration, indicating progress in understanding the topic.
GoodTimes
Messages
8
Reaction score
0
Hello, I am new here, glad I found this Forum :0)
I am german and english is not my mother-tongue, so sorry in advance for any language-related mistakes

1. Homework Statement

i don't know how to make a vector arrow on a letter so I declare a v for vector as the following:
[PLAIN]http://upload.wikimedia.org/math/8/8/b/88b537be1c5870ce272a4dc9515bada9.png=vr(t) A particle moves on a space-curve, described by following position-vector:

[PLAIN]http://upload.wikimedia.org/math/8/8/b/88b537be1c5870ce272a4dc9515bada9.png= (t²+t)vex + (3t-2)vey + (2t³-4t²)vez

Homework Equations

The Attempt at a Solution



I think I might differentiate each part, but I am not sure,

I would do it like this

vr'(t)= (2t)vex + (3)vey + (6t² - 8t)vez
so this would be the velocityand vr''(t) = (2)vex + (0)vey + (12t - 8)vez
would be the acceleration
 
Last edited by a moderator:
Physics news on Phys.org
Your problem statement does not include a question. Are you simply interested in finding the velocity and acceleration?

The basis vectors (in Cartesian coordinates) are constant so you can just differentiate the prefactors as you have done.

Edit: You can write mathematical notation on Physics Forums using LaTeX commands. Here is a very brief introduction. You can also press Reply on my post and you can see how I have done the following:
$$
\vec v(t) = \vec r'(t) = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z
$$
 
Orodruin said:
Your problem statement does not include a question. Are you simply interested in finding the velocity and acceleration?

The basis vectors (in Cartesian coordinates) are constant so you can just differentiate the prefactors as you have done.

Edit: You can write mathematical notation on Physics Forums using LaTeX commands. Here is a very brief introduction. You can also press Reply on my post and you can see how I have done the following:
$$
\vec r(t)
$$

I am sorry, yes correct, my task is to find the velocity and the acceleration.
does your post mean that the two equations I've done are correct? so i could simply put in a t, for example t=2 and calculate it
 
The principle is correct, but one of your derivatives is not. I suggest you double check your computations (alternatively that you typed it in correctly, it may be a typo as well).
 
ahhh it is

$$
\vec r'(t) = (2t+1)\vec e_x + (3)\vec e_y + (6t²-8t)\vec e_z
$$
Orodruin said:
Your problem statement does not include a question. Are you simply interested in finding the velocity and acceleration?

The basis vectors (in Cartesian coordinates) are constant so you can just differentiate the prefactors as you have done.

Edit: You can write mathematical notation on Physics Forums using LaTeX commands. Here is a very brief introduction. You can also press Reply on my post and you can see how I have done the following:
$$
\vec v(t) = \vec r'(t) = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z
$$
 
Correct.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top