How Do You Solve for Time in the Equation d = 0.5(at^2)?

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Homework Help Overview

The discussion centers around solving the equation d = 0.5(at^2) for time (t) in terms of acceleration (a) and distance (d). This falls within the subject area of kinematics.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different forms of the equation and attempt to manipulate it to isolate t. Some express uncertainty about the starting point, while others suggest alternative representations of the equation.

Discussion Status

The discussion includes various attempts to derive t from the equation, with some participants providing different expressions and insights. There is no explicit consensus, but several lines of reasoning are being explored.

Contextual Notes

Participants are working under the assumption that they need to isolate t without additional information or constraints provided in the problem statement.

besthands
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1. d = 0.5 (at^2) solve for t in terms of a and d

Probably really simple but I have no clue where to start from...
 
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d=(at^2/2) is the same as your equation, and try and solve from that
 
d = 0.5(a)(t^2), maybe that will help.
 
Alright I think I got it. A= 2d/t^2, t = square root of 2d/a.
 

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