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Let's say there is an object flying around (in a straight line), which has a constant speed $v$ and zero acceleration.

In some moment I can apply a constant deceleration $a$ to that object and I need to find a distance $d$ which will be traveled until the object stops.

Based on this formulas: $v = at$, $d = \frac{1}{2} at^2$ I wrote this:

$$d = \frac{v^2}{2a} $$

My question, is that correct and if not, what is ?

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    $\begingroup$ Yes, correct.${}{}{}$ $\endgroup$ Nov 8, 2012 at 20:25

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You're correct. Another way to solve this would have been to rearrange the following equation: $$v_f^2 = v_o^2 + 2ax$$ Because $v_f=0$ we can rewrite: $$x = {-v_o^2 \over 2a}$$ We can ignore the negative sign because the acceleration is negative, and will therefore cancel it out.

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