Respuesta :

gmany

[tex]v_0+\dfrac{1}{2}at^2=d\qquad|\text{subtract }v_0\ \text{from both sides}\\\\\dfrac{1}{2}at^2=d-v_o\qquad|\text{multiply both sides by 2}\\\\at^2=2(d-v_0)\qquad|\text{divide both sides by }t^2\\\\\boxed{a=\dfrac{2(d-v_0)}{t^2}}[/tex]

sid071

Hey there!!

We have:

... [tex]d=+v_{0} \frac{1}{2}a t^{2}[/tex]

Now, we will have to solve for 'a'.

Subtract v₀ on both sides.

... d - v₀ = at²/2

Now, let's multiply both sides with 2.

... 2(d-v₀) = (at²/2)2

... 2(d-v₀)=at²

Now, we will have to divide both sides with t².

... 2(d-v₀)/t² = a

This is the final answer.

Hope my answer helps!!