tion:Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration, \[\frac{d}{dt}v(t)=a(t),\]we can take the indefinite integral of both sides, FINDING \[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]where C1 is a constant of integration. Since \[\int \frac{d}{dt}v(t)dt=v(t)\], the velocity is GIVEN by \[v(t)=\int a(t)dt+{C}_{1}.\]Similarly, the time derivative of the position function is the velocity function, \[\frac{d}{dt}x(t)=v(t).\]Thus, we can use the same mathematical MANIPULATIONS we just used and find \[x(t)=\int v(t)dt+{C}_{2},\]where C2 is a second constant of integration.We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find \[v(t)=\int adt+{C}_{1}=at+{C}_{1}.\]If the INITIAL velocity is v(0) = v0, then \[{v}_{0}=0+{C}_{1}.\]Then, C1 = v0 and \[v(t)={v}_{0}+at,\]which is (Equation). Substituting this expression into (Figure) gives \[x(t)=\int ({v}_{0}+at)dt+{C}_{2}.\]Doing the integration, we find \[x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}.\]If x(0) = x0, we have \[{x}_{0}=0+0+{C}_{2};\]so, C2 = x0. Substituting BACK into the equation for x(t), we finally have \[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\] by using formula do ans