Solving a Rectilinear Motion Problem

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 \[ s=s(t) ={\frac{4}{3}}t^{3}-{\frac{3}{2}}t^{2}+4t+1 \]