Relation between velocity, position, and acceleration

The relation between velocity \( v \) and position \( x \), and acceleration \( a \) indicate as follows. \begin{eqnarray} x &=& \frac{1}{2}a t^2 + v_0 t \ \ \ \ \ \ \ \ \ \ (1) \\ v &=& at + v_0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{eqnarray} Here, \( v_0 \) shows the initial velocity. In university physics, most of the time, the differential calculus and integral calculus are often used to express the relationship \(v\) and \(x\). You may notice by intuition, the temporal change of velocity, position, and acceleration calculate from differentiating or integrate each of them. In other words, the distance can be determined by multiplying the velocity by a time. The amount of temporal change of distance equals velocity, so the differential value of the range will be velocity. \begin{eqnarray} v = \frac{dx}{dt}\ \ \ \ \ \ \ \ \ \ (3) \end{eqnarray} In contrast, the integrated value of velocity will be distance. \begin{eqnarray} x = \int v\ \mathrm{dt} = vt + x_0\ \ \ \ \ \ \ \ \ \ (4) \end{eqnarray} Here, \( x_0 \) shows the integral constant and called the initial position. The same thing can say about velocity and acceleration. The acceleration expresses the temporal change of velocity. \begin{eqnarray} a = \frac{dv}{dt} \ \ \ \ \ \ \ \ \ \ (5) \end{eqnarray} The solution is calculated by integrating Equation (5) and derive an Equation (2). \begin{eqnarray} v = \int a\ \mathrm{dt} = at + v_0\ \ \ \ \ \ \ \ \ \ (6) \end{eqnarray} where \( v_0 \) is the initial velocity. By substituting Equation (6) to (2), we get the following equation; \begin{eqnarray} x = \int( at + v_0 )\ \mathrm{dt} = \frac{1}{2} at^2 + v_0 t + x_0\ \ \ \ \ \ \ \ \ \ (7) \end{eqnarray} and if the initial position regarded as \( x_0=0 \) leads the Equation (1). Finally, velocity, position, and acceleration closely connected. The acceleration is second order differential of position. \begin{eqnarray} a = \frac{d^2x}{dt^2}\ \ \ \ \ \ \ \ \ \ \ (8) \end{eqnarray}