Formula: Uniformly Accelerated Motion Position Initial position Velocity
$$x(t) ~=~ x_0 ~+~ \class{blue}{v_0} \, t ~+~ \frac{1}{2} \, \class{red}{a} \, t^2$$
$$x(t) ~=~ x_0 ~+~ \class{blue}{v_0} \, t ~+~ \frac{1}{2} \, \class{red}{a} \, t^2$$
$$x_0 ~=~ x ~-~ \class{blue}{v_0} \, t ~-~ \frac{1}{2} \, \class{red}{a} \, t^2$$
$$\class{blue}{v_0} ~=~ \frac{1}{t} \, \left( x - x_0 - \frac{\class{red}{a}\, t^2}{2} \right)$$
$$t ~=~ -\frac{\class{blue}{v_0}}{\class{red}{a}} \pm \sqrt{ \left( \frac{\class{blue}{v_0}}{\class{red}{a}} \right)^2 ~-~ \frac{2(x_0-x)}{\class{red}{a}} }$$
$$\class{red}{a} ~=~ \frac{2}{t^2} \, \left( x - x_0 - \class{blue}{v_0} \, t \right)$$
Position
$$ x(t) $$ Unit $$ \mathrm{m} $$
Current position \(x(t)\) of an accelerated object at time \(t\) on the \(x\) axis.
Initial position
$$ x_0 $$ Unit $$ \mathrm{m} $$
Position of the object at the time \(t = 0\). It is usual to set the initial position \(x_0 = 0\). Then the position \(x(t)\) corresponds to the distance covered by the body after the time \(t\).
Initial velocity
$$ \class{blue}{v_0} $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$
Velocity with which the object starts the acceleration process. If the object starts from rest, then the initial velocity is \( v_0 = 0\).
Time
$$ t $$ Unit $$ \mathrm{s} $$
This is the time during which the object accelerates.
Acceleration
$$ \class{red}{\boldsymbol a} $$ Unit $$ \frac{\mathrm m}{\mathrm{s}^2} $$
Uniform acceleration of the object. That means: Every second the current velocity of the body increases by the value \(a\).