How would you rework this equation $v_{f}^{2} = v_{i}^{2} - 2 \cdot a \cdot d$ $v_f^2= v_i^2 - 2ad$ so that it solves for $a$ $a$ ?

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Answer 1 · 2015-07-20T22:33:39

You simply isolate $a$ $a$ on one side of the equation.

If you start from this equation

$v_{f}^{2} = v_{i}^{2} - 2 \cdot a \cdot d$ $v_f^2 = v_i^2 - 2 * a * d$

you can solve for the acceleration of the object, $a$ $a$ , by

$v_f^2 - v_i^2 = cancel(v_i^2) - cancel(v_i^2) - 2 * a * d$

$v_f^2 - v_i^2 = -2 * a * d$

$(v_f^2 - v_i^2)/((-2 * d)) = a * cancel((-2 * d))/cancel((-2 * d))$

This is equivalent to

$a = -(v_f^2-v_i^2)/(2d)$ , or

$a = color(green)((v_i^2 - v_f^2)/(2d))$

How would you rework this equation v_f^2= v_i^2 - 2*a*dv2f=v2i−2⋅a⋅dv_f^2= v_i^2 - 2*a*d so that it solves for aaa?