Solve for t x=x_0+v_0t+1/2at^2

$x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$

Rewrite the equation as $x_{0} + v_{0} t + \frac{1}{2} \cdot (a t^{2}) = x$ .

$x_{0} + v_{0} t + \frac{1}{2} \cdot (a t^{2}) = x$

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

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

$x_{0} + v_{0} t + \frac{a}{2} t^{2} = x$

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

$x_{0} + v_{0} t + \frac{a t^{2}}{2} = x$

Subtract $x$ from both sides of the equation.

$x_{0} + v_{0} t + \frac{a t^{2}}{2} - x = 0$

Multiply through by the least common denominator $2$ , then simplify.

Apply the distributive property.

$2 x_{0} + 2 (v_{0} t) + 2 (\frac{a t^{2}}{2}) + 2 (- x) = 0$

Simplify.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 x_{0} + 2 v_{0} t + 2 (\frac{a t^{2}}{2}) + 2 (- x) = 0$

Rewrite the expression.

$2 x_{0} + 2 v_{0} t + a t^{2} + 2 (- x) = 0$

Multiply $- 1$ by $2$ .

$2 x_{0} + 2 v_{0} t + a t^{2} - 2 x = 0$

Move $v_{0}$ .

$2 x_{0} + 2 t v_{0} + a t^{2} - 2 x = 0$

Move $2 x_{0}$ .

$2 t v_{0} + a t^{2} + 2 x_{0} - 2 x = 0$

Reorder $2 t v_{0}$ and $a t^{2}$ .

$a t^{2} + 2 t v_{0} + 2 x_{0} - 2 x = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = a$ , $b = 2 v_{0}$ , and $c = 2 x_{0} - 2 x$ into the quadratic formula and solve for $t$ .

$\frac{- 2 v_{0} \pm \sqrt{{(2 v_{0})}^{2} - 4 \cdot (a \cdot (2 x_{0} - 2 x))}}{2 a}$

Simplify.

Simplify the numerator.

Add parentheses.

$t = \frac{- 2 v_{0} \pm \sqrt{{(2 v_{0})}^{2} - 4 \cdot (a \cdot (2 x_{0} - 2 x))}}{2 a}$

Let $u = a \cdot (2 x_{0} - 2 x)$ . Substitute $u$ for all occurrences of $a \cdot (2 x_{0} - 2 x)$ .

Apply the product rule to $2 v_{0}$ .

$t = \frac{- 2 v_{0} \pm \sqrt{2^{2} {v_{0}}^{2} - 4 \cdot u}}{2 a}$

Raise $2$ to the power of $2$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 {v_{0}}^{2} - 4 u}}{2 a}$

Factor $4$ out of $4 {v_{0}}^{2} - 4 u$ .

Factor $4$ out of $4 {v_{0}}^{2}$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 {v_{0}}^{2} - 4 u}}{2 a}$

Factor $4$ out of $- 4 u$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 {v_{0}}^{2} + 4 (- u)}}{2 a}$

Factor $4$ out of $4 {v_{0}}^{2} + 4 (- u)$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - u)}}{2 a}$

Replace all occurrences of $u$ with $a \cdot (2 x_{0} - 2 x)$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - (a \cdot (2 x_{0} - 2 x)))}}{2 a}$

Simplify each term.

Apply the distributive property.

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - (a (2 x_{0}) + a (- 2 x)))}}{2 a}$

Rewrite using the commutative property of multiplication.

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - (2 a x_{0} + a (- 2 x)))}}{2 a}$

Rewrite using the commutative property of multiplication.

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - (2 a x_{0} - 2 a x))}}{2 a}$

Apply the distributive property.

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - (2 a x_{0}) - (- 2 a x))}}{2 a}$

Multiply $2$ by $- 1$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - 2 (a x_{0}) - (- 2 a x))}}{2 a}$

Multiply $- 2$ by $- 1$ .

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - 2 (a x_{0}) + 2 (a x))}}{2 a}$

Remove parentheses.

$t = \frac{- 2 v_{0} \pm \sqrt{4 ({v_{0}}^{2} - 2 a x_{0} + 2 a x)}}{2 a}$

Rewrite $4$ as $2^{2}$ .

$t = \frac{- 2 v_{0} \pm \sqrt{2^{2} ({v_{0}}^{2} - 2 a x_{0} + 2 a x)}}{2 a}$

Pull terms out from under the radical.

$t = \frac{- 2 v_{0} \pm 2 \sqrt{{v_{0}}^{2} - 2 a x_{0} + 2 a x}}{2 a}$

Simplify $\frac{- 2 v_{0} \pm 2 \sqrt{{v_{0}}^{2} - 2 a x_{0} + 2 a x}}{2 a}$ .

$t = \frac{- v_{0} \pm \sqrt{{v_{0}}^{2} - 2 a x_{0} + 2 a x}}{a}$

The final answer is the combination of both solutions.

$t = - \frac{v_{0} - \sqrt{{v_{0}}^{2} - 2 a x_{0} + 2 a x}}{a}$

$t = - \frac{v_{0} + \sqrt{{v_{0}}^{2} - 2 a x_{0} + 2 a x}}{a}$

Solve for t x=x_0+v_0t+1/2at^2

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