What is the Time Equation for Constant Acceleration?

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The discussion focuses on finding an equation for time as a function of initial velocity, distance, and constant acceleration. The user shares simulation results for distances of 25 and 50 meters, with specific initial conditions, but struggles to derive an accurate equation. Suggestions include using the kinematic equation s = v0t + 1/2at² and solving for time using the quadratic formula. The user confirms that acceleration is constant and seeks clarification on solving the quadratic without calculus. The conversation emphasizes the importance of understanding basic kinematic principles for accurate calculations.
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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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