Acceleration Engineering First Semester

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Discussion Overview

The discussion revolves around a physics problem concerning the position function of a mass point, specifically analyzing its velocity and acceleration at a given time. Participants explore the implications of the function and seek clarification on the correct approach to derive speed and acceleration.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents the position function x(t) = 4(m·s)t - 1 + 10m and asks about the direction of speed, constancy of speed, sign changes in velocity, and direction of acceleration at t = 1 s.
  • Another participant reminds the original poster to show their best attempt at solving the problem, referencing forum guidelines.
  • A participant suggests that the mass point is at a height of 10m and proposes that differentiation of the position function is necessary to find speed at t = 1 s.
  • A later reply questions the correctness of the initial formula and suggests a different representation of the position function, indicating that velocity is derived from the position by differentiation.
  • Another participant acknowledges a mistake in the formula and provides a corrected version, while expressing gratitude for the previous response.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct formulation of the position function, and there is uncertainty regarding the steps needed to solve the problem. Multiple viewpoints on the interpretation of the function and its implications remain present.

Contextual Notes

There are unresolved issues regarding the correct interpretation of the position function and the necessary steps for differentiation. The discussion reflects varying assumptions about the formula's structure.

germanmath
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Homework Statement
Acceleration
Relevant Equations
Engineering, biomedical, University
The function x (t) = 4 (m · s) t − 1 + 10 m describes the position of a mass point. In which direction does the speed point at time t = 1 s? Is the amount of speed constant or not? does the velocity function change its sign for t> 0 at some point? And in which direction does the acceleration point at time t = 1 s?

could someone help me please with Solving this task?
 
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Oh of course, I am sorry.

I would say that the mass point positions is in the height of 10m...
To get the speed I assume that I have to do the derivitation and then I have to calculate the speed for the value of t=1. is it right?
 
germanmath said:
Oh of course, I am sorry.

I would say that the mass point positions is in the height of 10m...
To get the speed I assume that I have to do the derivitation and then I have to calculate the speed for the value of t=1. is it right?
Is the formula supposed to be? $$x(t) = (4 m/s)(t-1s) + 10m$$ In any case, you get the velocity from the position by differentiation.
 
No, I am sorry I didnt noticed that it is incorrectly. The formula is supposed to be: x(t) =4(m*s)(t^-1) +10meter

okey thanks a lot for your answer :)
 

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