Solve for a d=vt+1/2at^2

$d = v t + \frac{1}{2} a t^{2}$

Rewrite the equation as $v t + \frac{1}{2} \cdot (a t^{2}) = d$ .

$v t + \frac{1}{2} \cdot (a t^{2}) = d$

Multiply $\frac{1}{2} (a t^{2})$ .

Combine $a$ and $\frac{1}{2}$ .

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

Combine $\frac{a}{2}$ and $t^{2}$ .

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

Subtract $v t$ from both sides of the equation.

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

Multiply both sides of the equation by $2$ .

$2 \frac{a t^{2}}{2} = 2 (d - v t)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{a t^{2}}{2} = 2 (d - v t)$

Rewrite the expression.

$a t^{2} = 2 (d - v t)$

Simplify the right side.

Simplify $2 (d - v t)$ .

Apply the distributive property.

$a t^{2} = 2 d + 2 (- v t)$

Multiply $- 1$ by $2$ .

$a t^{2} = 2 d - 2 v t$

Divide each term in $a t^{2} = 2 d - 2 v t$ by $t^{2}$ and simplify.

Divide each term in $a t^{2} = 2 d - 2 v t$ by $t^{2}$ .

$\frac{a t^{2}}{t^{2}} = \frac{2 d}{t^{2}} + \frac{- 2 v t}{t^{2}}$

Simplify the left side.

Cancel the common factor of $t^{2}$ .

Cancel the common factor.

$\frac{a t^{2}}{t^{2}} = \frac{2 d}{t^{2}} + \frac{- 2 v t}{t^{2}}$

Divide $a$ by $1$ .

$a = \frac{2 d}{t^{2}} + \frac{- 2 v t}{t^{2}}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $t$ and $t^{2}$ .

Factor $t$ out of $- 2 v t$ .

$a = \frac{2 d}{t^{2}} + \frac{t (- 2 v)}{t^{2}}$

Cancel the common factors.

Factor $t$ out of $t^{2}$ .

$a = \frac{2 d}{t^{2}} + \frac{t (- 2 v)}{t \cdot t}$

Cancel the common factor.

$a = \frac{2 d}{t^{2}} + \frac{t (- 2 v)}{t \cdot t}$

Rewrite the expression.

$a = \frac{2 d}{t^{2}} + \frac{- 2 v}{t}$

Move the negative in front of the fraction.

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

Solve for a d=vt+1/2at^2

Related Questions

Support LCMGCF.com by disabling your adblocker.