Question

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

Answer 1

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$