Finding distance with known constant acceleration

Click For Summary

Homework Help Overview

The discussion revolves around calculating the distance traveled by a person coming to a complete stop under constant acceleration, specifically at an acceleration of 60g over a time period of 36 milliseconds.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the distance, converting acceleration and time into usable values. Some participants question the interpretation of the acceleration value and the signs used in the equations.

Discussion Status

Participants are actively engaging with the problem, with one offering a calculation attempt and others providing feedback on potential misunderstandings regarding variable definitions and signs. There is no explicit consensus yet, but guidance has been offered regarding the correct identification of initial and final velocities.

Contextual Notes

There is a noted confusion regarding the acceleration value expressed in terms of g's, as well as the proper application of kinematic equations in the context of the problem.

tater08
Messages
29
Reaction score
0
1. How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g?



2. X=Xo+ volt and V1=Vo +at



3. I converted 60g into 588.6 m/s^2 and then plugged it into the V1 equation to Find V1 which turns out to be 21.1896 m/s (after converting 36 ms into 0.036 seconds). I then entered the 21 m/s into the X1 equation and X=0+21.1896 * 0.036. I keep getting x = 0.76m. I am not confident that I am doing this question right or what I should be doing but that is my logic.
 
Physics news on Phys.org
60g? Six, zero, g's?
 
yup 60 g's.
 
Last edited:
You've got v1 and v0 mixed up.

You are given v1; it is 0. You need to solve for v0. (and then make sure you've got your signs correct for both v and a.)
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
15K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K