When Does the Particle Reach Its Maximum Positive X Position?

  • Thread starter Thread starter bmadkins
  • Start date Start date
  • Tags Tags
    Displacement Time
Click For Summary
SUMMARY

The particle's position along the x-axis is defined by the equation x(t) = 100 t^2 - 8 t^3. To determine when the particle reaches its maximum positive x position, one must find the time at which its velocity, derived from the position function, equals zero. The length of the path covered by the particle between t = 0.0000 s and t = 10.833 s can be calculated using arc length formulas, while the displacement is the straight-line distance between the two time points.

PREREQUISITES
  • Understanding of calculus, specifically local maxima and minima
  • Familiarity with the concept of velocity as the derivative of position
  • Knowledge of arc length calculations in calculus
  • Basic principles of kinematics in physics
NEXT STEPS
  • Review Calculus I concepts on finding local maxima and minima
  • Learn how to derive velocity from a position function
  • Study arc length formulas and their applications
  • Explore kinematic equations for non-uniform motion
USEFUL FOR

Students studying physics, particularly those tackling problems involving motion along a line, as well as educators looking for teaching resources on calculus and kinematics.

bmadkins
Messages
1
Reaction score
0

Homework Statement


The position of a particle moving along the x-axis depends on time as x(t) = 100 t^2 - 8 t^3 where x is in m and t in s. At what time does the particle reach its maximum positive x position?

What is the length of the path covered by the particle between t = 0.0000 s and t = 10.833 s?

What is the displacement of the particle between t = 0.0000 s and t = 10.8333 s?


Homework Equations


x(t) = 100 t^2 - 8 t^3



The Attempt at a Solution


I haven't had any attempts that have led me anywhere, really. This is the first time I've encountered this type of problem.
 
Physics news on Phys.org
The length of the path would be the arc length between the two points.

The displacement would be the straight line distance between the two points.
 
Hello bmadkins,

Welcome to Physics Forums!
bmadkins said:
I haven't had any attempts that have led me anywhere, really. This is the first time I've encountered this type of problem.
The best way to do the first part of problem (finding the time at which the position is positive maximum) is to open up your Calculus I textbook or notes and review the section of how to find the local maximums and local minimums of a function.

But perhaps a more intuitive guide (which is the same thing applied to this problem), is to consider that at some moment in time, the particle is moving away from its original position. Eventually, it slows down, momentarily stops, and then goes back in the other direction toward the original position.

So at what point was it farthest away from the original position? At the instant it momentarily stopped -- at the instant it "turned around." At that point its instantaneous velocity is zero. So part of this problem is figuring out at what time (or times) the particle's velocity is zero.

So to start, you might want to ask yourself, "what is a particle's velocity (as a function of time) if you already know what its position is (also as a function of time)?"

(And I don't mean s = vt, since the particle is not moving at a constant velocity. And I don't mean v = at or s = ½at2 either, since the acceleration is not necessarily uniform. There is a more fundamental relationship between position and velocity.)
 

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K