Translating acceleration into distance traveled

AI Thread Summary
To calculate the distance traveled under constant acceleration from rest, the relevant equation is s = v₀t + (1/2)at², where v₀ is the initial velocity, a is acceleration, t is time, and s is distance. For a car accelerating at 2 m/s² for 20 seconds, the initial velocity (v₀) is 0, making the equation simplify to s = (1/2)(2)(20²), resulting in a distance of 400 meters. Another useful equation is v₁² - v₀² = 2as, which relates final velocity and distance. Understanding these equations allows for accurate calculations of distance traveled during uniformly accelerated motion. Mastery of these concepts is essential for algebra-based physics.
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Homework Statement



If something is moving at a constant acceleration (starting from 0 velocity) for a known period of time, how do you figure how far they've traveled? Eg. a car is accelerating at 2 m/s^2 for 20 seconds.

I'm taking an algebra based physics course.

Homework Equations



I do not know which equation are relevant.

The Attempt at a Solution



I know I can multiply time by acceleration and get a final velocity. for example the final velocity is 40 m/s in my example. I don't know how to factor position into it.
 
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Look up Newton's equations of motion

I can tell you how to derive them if you wish.
 
Hello,

I think the equations you need are those we use for uniform motion, ie when the acceleration is constant.

Assuming all the motion is one direction only the distance traveled will be s=v_{0}t + \frac{at^{2}}{2}

where v_{0}=initial\ velocity\; a=acceleration\; t=time\ elapsed\; and\ s=distance\ traveled

Also v_{1}^{2}-v_{0}^{2}=2as
 
Vuldoraq said:
Hello,

I think the equations you need are those we use for uniform motion, ie when the acceleration is constant.

Assuming all the motion is one direction only the distance traveled will be s=v_{0}t + \frac{at^{2}}{2}

where v_{0}=initial\ velocity\; a=acceleration\; t=time\ elapsed\; and\ s=distance\ traveled

Also v_{1}^{2}-v_{0}^{2}=2as

Thanks that works
 
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