What is the Time Equation for Constant Acceleration?

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Homework Help Overview

The discussion revolves around finding an equation for time as a function of initial velocity, distance, and acceleration in the context of constant acceleration. The original poster shares simulation results for specific distances and seeks assistance in deriving an appropriate equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and explore different formulations for time. There are attempts to derive time from known variables, and questions arise about the validity of certain equations and the rounding of results.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on relevant equations and methods for solving quadratic equations. There is acknowledgment of the need to clarify assumptions about constant acceleration, and the original poster indicates a willingness to follow up on findings.

Contextual Notes

There is mention of specific simulation data and initial conditions, as well as a note that the original poster is new to physics at a college level, which may influence their understanding of the concepts discussed.

kriegera
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Hi All - I am trying to find an equation that gives the time as a function of initial velocity, distance traveled, and specified acceleration. That is, t = some function of v(0), d, and a.

We ran a simulation problem for distances 25 and 50 meters and my times were - 4.099 and 6.141 respectively for initial velocity of 2.00 and inital acceleration of 2.00. i can't come up with any equations that fit my data. any input please?

Thank you!
Ashley
 
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So would it be:
t=SQRT(2s-v(o)/a

s=distance
v(o)=inital velocity

I've tried this but it doesn't give me the exact anser - is it ok to round up or down? Initally I found final velocity and then plugged it in but the challenge is to use one equation. Input?

Thank you!
 
kriegera said:
So would it be:
t=SQRT(2s-v(o)/a
The starting equation would be:
s = v0t + 1/2at2

To find t you'll need to solve the quadratic.
 
Doc Al said:
The starting equation would be:
s = v0t + 1/2at2

To find t you'll need to solve the quadratic.

Thank you for this tip. I am not very "up" on my physics - this is my first introduction to it on a college level. What are the rules for solving for the quadratic? Does that involve derivitives? Thank you!
 
kriegera said:
What are the rules for solving for the quadratic? Does that involve derivitives?
No need for calculus, just the 'quadratic formula'. (Google it!)

Note that I am assuming that your setup is one where the acceleration is constant. (You didn't actually state that.)
 
Note that I am assuming that your setup is one where the acceleration is constant. (You didn't actually state that.)[/QUOTE said:
Yes - acceleration is constant. I'll let you know what I find out. Thanks.
 

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