Find the distance $s$ of an object from the origin at time $t$ if its acceleration $a$ is $$a(t)=8t-3$$

and the initial conditions are $v_{0}=v(0)=4$ and $s_{0}=s(0)=1.$

Solution First we solve the differential equation $\dfrac{{d\kern-1ptv}}{dt} =a(t)=8t-3$ and use the initial condition $v_{0}=v( 0) =4$. $$\begin{eqnarray*} v(t) &=& 4t^{2}-3t+C_{1} \\[3pt] {v_{0}} &=& v( 0) =4( 0) ^{2}-3( 0) + C_{1}&=&4 \\[0pt] &&\qquad\quad\hspace{3.9pc}C_{1}&=&4 \end{eqnarray*}$$

The velocity of the object at time $t$ is $v(t)=4t^{2}-3t+4$.

The distance $s$ of the object at time $t$ satisfies the differential equation $$\frac{ds}{dt}=v(t)=4t^{2}-3t+4$$

Then $${s(t)={\frac{4}{3}}t^{3}-{\frac{3}{2}}t^{2}+4t+C_{2}}$$

Using the initial condition, $s_{0}=s(0)=1$, we have $$\begin{eqnarray*} {s_{0}=s(0)=0-0+0+C_{2}} &=&1 \\ {C_{2}} &=&1 \end{eqnarray*}$$

The distance $s$ of the object from the origin at any time $t$ is $$s=s(t) ={\frac{4}{3}}t^{3}-{\frac{3}{2}}t^{2}+4t+1$$