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

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

Step 1

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

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

Step 2

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 $t^{2}$ .

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

Step 3

Subtract $\frac{a t^{2}}{2}$ from both sides of the equation.

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

Step 4

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

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

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

Simplify the left side.

Cancel the common factor of $t$ .

Cancel the common factor.

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

Divide $v$ by $1$ .

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

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $t$ .

Move the leading negative in $- \frac{a t^{2}}{2}$ into the numerator.

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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