Q. 49 - Calculus

Question

Consider a toy car that moves back and forth on a long, straight track for $4$ minutes in such a way that the function $s\left(t\right)=96t-84t^2+28t^3-3t^4$describes how far the car is to the right of the starting point, in centimeters, after $t$ minutes. 

       Within the domain of your model, when is the toy car moving fastest to the right? When is the toy car moving fastest to the left? 

Step-by-Step Solution

Answer

Ans:  The toy car is the fastest at $t=0$ minutes and at the right endpoint of the model such as $t=5$

Step 1. Given information.

given expression,  $s\left(t\right)=96t-84t^2+28t^3-3t^4$

Step 2. The objective is to determine the time when the toy car is the fastest to the right and to the left.

Finding its derivative, 

     $\begin{array}{r}{s}^{\mathrm{\prime }}\left(t\right)=v\left(t\right)=96-168t+84t^2-12t^3\\ s^{\prime \prime }\left(t\right)=v^{\mathrm{\prime }}\left(t\right)=-168+168t-36t^2\\ s^{-\mathrm{\prime }}\left(t\right)=v^{\mathrm{\prime }}\left(t\right)=168-72t\\ s^{-}\left(t\right)=-72 \end{array}$

Since, $s^{-}\left(t\right)<0$ so there exist a maximum.

Step 3. The maximum exists at,

   $\begin{array}{r}{s}^{\prime \prime }\left(t\right)=0\\ -168+168t-36t^2=0\\ t=\frac{7-\sqrt{7}}{3},\frac{7+\sqrt{7}}{3}\end{array}$

Putting  $t=\frac{7-\sqrt{7}}{3},\frac{7+\sqrt{7}}{3}.0,4$ in $v^{\mathrm{\prime }}\left(t\right)=168-72t$ the maximum occurs when $t=0$

      $v^{\mathrm{\prime }}\left(0\right)=168-72\left(0\right)=168$

Therefore,  the toy car is the fastest at $t=0$ minutes and at the right endpoint of the model such as $t=5$