# How do you solve for a: in y= (at²)÷2 + vt?

Mar 23, 2018

$a = \frac{2 \left(y - v t\right)}{t} ^ 2$

#### Explanation:

The function $y = \left(a {t}^{2}\right) \div 2 + v t$ represents distance covered by an object starting with an initial velocity $v$ with an accelaration $a$, in time $t$.

As $y = \left(a {t}^{2}\right) \div 2 + v t$

$\left(a {t}^{2}\right) \div 2 = y - v t$

i.e. $a {t}^{2} = 2 \left(y - v t\right)$

and $a = \frac{2 \left(y - v t\right)}{t} ^ 2$

Mar 23, 2018

a=(−2tv+2y)/t^2

#### Explanation:

Step 1: Simplify

$y = \frac{a {t}^{2}}{2} + v t$
Step 2: Flip the equation.

1/2 at^2+tv=y
Step 3: Add -tv to both sides.

1/2 at^2+tv+color(red)(−tv)=y+color(red)(−tv)
1/2at2=−tv+y

Step 4: Divide both sides by ${t}^{2} / 2$.
(1/2at^2)/color(red)(t^2/2)=(−tv+y)/color(red)(t^2/2)
a=(−2tv+2y)/(t^2)