How do I find the speed of an object with a given position function?

  • Thread starter Thread starter Math10
  • Start date Start date
  • Tags Tags
    Speed
Click For Summary
SUMMARY

The speed of an object defined by the position function r(t) = e^(t) is determined by calculating the magnitude of the velocity vector v(t) = e^(t). The correct speed is confirmed to be |v(t)| = sqrt(3)*e^(t). This conclusion is reached by applying the formula for the magnitude of a vector, which is essential in vector calculus.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with the concept of velocity and speed
  • Knowledge of exponential functions and trigonometric functions
  • Ability to compute the magnitude of a vector
NEXT STEPS
  • Study the calculation of vector magnitudes in depth
  • Explore the properties of exponential and trigonometric functions
  • Learn about the applications of velocity and speed in physics
  • Investigate the use of parametric equations in motion analysis
USEFUL FOR

Students in physics or mathematics, particularly those studying calculus or vector analysis, will benefit from this discussion. It is also useful for educators teaching concepts of motion and speed in a mathematical context.

Math10
Messages
301
Reaction score
0

Homework Statement


Find the speed of an object with r(t)=e^(t)<cos(t), sin(t), 1>.

Homework Equations


None.[/B]

The Attempt at a Solution


v(t)=e^(t)<cos(t)-sin(t), cos(t)+sin(t), 1>
I know that the speed is abs(v(t)) and the answer in the book is sqrt(3)*e^(t). So how do I find the speed?
 
Physics news on Phys.org
Math10 said:

Homework Statement


Find the speed of an object with r(t)=e^(t)<cos(t), sin(t), 1>.

Homework Equations


None.[/B]

The Attempt at a Solution


v(t)=e^(t)<cos(t)-sin(t), cos(t)+sin(t), 1>
I know that the speed is abs(v(t)) and the answer in the book is sqrt(3)*e^(t). So how do I find the speed?
The speed is the magnitude of your vector for v(t). How do you usually find the magnitude of a vector?
 
Never mind, I got it! Thanks for helping.
 

Similar threads

Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K