Register to reply

Displacement vs Time problem

by bmadkins
Tags: displacement, time
Share this thread:
bmadkins
#1
Sep1-11, 04:22 PM
P: 1
1. The problem statement, all variables and given/known data
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?


2. Relevant equations
x(t) = 100 t^2 - 8 t^3



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.
Phys.Org News Partner Science news on Phys.org
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage
rock.freak667
#2
Sep1-11, 06:11 PM
HW Helper
P: 6,207
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.
collinsmark
#3
Sep1-11, 06:19 PM
HW Helper
PF Gold
collinsmark's Avatar
P: 1,920
Hello bmadkins,

Welcome to Physics Forums!
Quote Quote by bmadkins View Post
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 = Żat2 either, since the acceleration is not necessarily uniform. There is a more fundamental relationship between position and velocity.)


Register to reply

Related Discussions
Time- Displacement diagram Introductory Physics Homework 3
How do I calculate speed in a velocity-time graph? Displacement-time graph? Introductory Physics Homework 1
Velocity as function of Displacement to Displacement as function of Time... General Physics 3
Time and Displacement Introductory Physics Homework 10
Solving for time and displacement... Introductory Physics Homework 1