Find the time taken by the truck to move 1 metre

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Homework Help Overview

The discussion revolves around a kinematics problem involving the time taken by a truck to move a distance of 1 meter. Participants are examining the equations related to motion under constant acceleration and questioning the accuracy of a textbook reference.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the correct formula to use for calculating time, with some suggesting that the textbook contains an error. There is a focus on the relationship between distance, acceleration, and time in the context of kinematics.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the equations. Some have provided alternative formulations for time calculation, while others are questioning the assumptions made in the problem setup.

Contextual Notes

There is mention of specific values and conditions, such as initial velocity being zero and the acceleration value used in the calculations. The discussion also references a textbook, indicating a reliance on external material for guidance.

chwala
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Homework Statement
See attached.
Relevant Equations
kinematics
Unless i am missing something; there is an error with the textbook

1707127143855.png


It ought to be ,

## t= \sqrt {\dfrac{1}{0.04}}= \sqrt 25 = 5## seconds.
 
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Why ?
 
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chwala said:
Homework Statement: See attached.
Relevant Equations: kinematics

Unless i am missing something; there is an error with the textbook

View attachment 339804

It ought to be ,

## t= \sqrt {\dfrac{1}{0.04}}= \sqrt 25 = 5## seconds.


If u = 0 then s = \tfrac12 at^2 \quad \Rightarrow \quad t = \sqrt{\frac{2s}{a}} as the textbook has, not t = \sqrt{s/a} as you are suggesting.
 
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pasmith said:
If u = 0 then s = \tfrac12 at^2 \quad \Rightarrow \quad t = \sqrt{\frac{2s}{a}} as the textbook has, not t = \sqrt{s/a} as you are suggesting.
Let me take a break! Cheers @pasmith
 
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