Quick Acceleration Formula: Solving for Time, Distance, and Acceleration

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Homework Help Overview

The discussion revolves around the application of kinematic equations to solve for time given acceleration and distance. The original poster seeks clarification on the appropriate formula to use, particularly in a scenario involving constant acceleration over a significant distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the kinematic equation for distance and how it applies when initial velocity is zero. There is a focus on deriving time from the equation and verifying its correctness. Some participants also explore implications of acceleration over astronomical distances and question the validity of Newtonian physics at high speeds.

Discussion Status

Participants have provided various interpretations of the kinematic equations and engaged in a discussion about the implications of acceleration in a theoretical context. There is acknowledgment of the need to consider relativistic effects when discussing speeds approaching that of light, indicating a productive exploration of the topic.

Contextual Notes

Some participants express uncertainty about their mathematical skills and the application of physics concepts, while others question the assumptions regarding the feasibility of traveling at relativistic speeds.

Kristopher
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My physics is so rusty and I'm at work so I can't dig out my books but I have a quick question.

What is the formula to solve for time if one knows acceleration and distance?

Example: if one accelerates 10m/s/s for 1000km, how long does it take to cover the distance?

Anyone remember acceleration equations or have them handy?

Thanks

PS this isn't homework I'm trying to explain something to a co-worker and I'm just really rusty.
 
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[tex]d = Vi + \frac{1}{2}*a*t^2[/tex]

However, in your case [tex]Vi = 0[/tex]
 
imranq said:
[tex]d = Vi + \frac{1}{2}*a*t^2[/tex]

However, in your case [tex]Vi = 0[/tex]

That's not the right general formula! If in the initial problem the car starts from rest it will not matter (since vi=0 in this case), but the correct formula is [tex]d=v_it+\frac{1}{2}at^2[/tex] Where d denotes the distance, vi the initial velocity, t the time, and a the acceleration. Note that you will need to convert the distance into metres.
 
Arg! I can't believe I forgot that, yeah it is Vi*t + .5at^2
 
My math is a little rusty too, but that would work out (since Vi is 0, Vi*t would equal 0) to be t= sqrt of 2d/a, yes??
 
Kristopher said:
My math is a little rusty too, but that would work out (since Vi is 0, Vi*t would equal 0) to be t= sqrt of 2d/a, yes??

Yes, that's correct.
 
So then if one were to create a thruster that could continually supply the force needed for 10m/s/s of acceleration or just over 1g. in 2 years one would have traveled 2.11 light years and have exceeded the speed of light... right?

ly=9,460,073,432,260,800 meters
multiply that by 2.11 (half the distance to alpha centauri)
d=19,960,754,942.070.288
a=10

work that out and t=63,183,471s which is about 2 years.

obviously if one travels 2.11 light years in 2 years, one exceeded the speed of light (it that is possible)

but if it is and if one could create some nuclear drive or other perpetual thruster, one could accelerate at just over 1g halfway to alpha centauri, turn around and decelerate at just over -1g, making it to alpha centauri, which is 4.22light years away, in 4 years with "gravity" the entire time except during the rotation.

Right?
 
One cannot exceed the speed of light. When traveling at large speeds, the equation I provided is no longer valid as one must take into account relativistic effects.
 
If Newtonian physics applied at these speeds, would my math be correct?
 

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