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copied!<p>This is a simple kinematics problem.</p> <p>At some time t, the velocity (v) of an object under constant acceleration is described by:</p> <pre><code>v = v0 + at </code></pre> <p>Where v0 is the initial velocity and a is the acceleration. In your case, the final velocity is zero (the object is stopped) so we can solve for t:</p> <pre><code>t = -v0/a </code></pre> <p>To find the total difference traveled, we take the integral of the velocity (the first equation) over time. I haven't done an integral in years, but I'm pretty sure this one works out to:</p> <pre><code>d = v0t + 1/2 * at^2 </code></pre> <p>We can substitute in the equation for t we developed ealier:</p> <pre><code>d = v0^2/a + 1/2 * v0^2 / a </code></pre> <p>And the solve for v0:</p> <pre><code>v0 = sqrt(-2ad) </code></pre> <p>Or, in a more programming-language format:</p> <pre><code>initialVelocity = sqrt( -2 * acceleration * distance ); </code></pre> <p>The acceleration in this case is negative (the object is slowing down), and I'm assuming that it's constant, otherwise this gets more complicated.</p> <p>If you want to use this inside a loop with a finite number of steps, you'll need to be a little careful. Each iteration of the loop represents a period of time. The object will move an amount equal to the <em>average</em> velocity times the length of time. A sample loop with the length of time of an iteration equal to 1 would look something like this:</p> <pre><code>position = 0; currentVelocity = initialVelocity; while( currentVelocity > 0 ) { averageVelocity = currentVelocity + (acceleration / 2); position = position + averageVelocity; currentVelocity += acceleration; } </code></pre>

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