Q. 48 - 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.

      When is the toy car farthest away from the starting point? Does it ever return to the starting point? For how long does this model make sense? 

Step-by-Step Solution

Answer

Ans:   The toy car is the farthest at $t=2$

Step 1. Given information.

given, 

      $s\left(t\right)=96t-84t^2+28t^3-3t^4$$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 from the starting point.

Finding its derivative,

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

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

 

Step 3. The maximum exists at,

 $\begin{array}{r}{s}^{\mathrm{\prime }}\left(t\right)=0\\ 96-168t+84t^2-12t^3=0\\ 2^5\cdot 3-\left(2^3\cdot 3\cdot 7\right)t+\left(2^2\cdot 3\cdot 7\right)t^2-\left(2^2\cdot 3\right)t^3\\ t=2,4,1\end{array}$

Putting $t=2,4,1$ in $s^{\ast }\left(t\right)=-168+168t-36t^2$

   $\begin{array}{r}{s}^{\ast }\left(2\right)=-168+168\left(2\right)-36(2{)}^2=24\\ s^{\ast }\left(4\right)=-168+168\left(4\right)-36(4{)}^2=-72\\ s^{\ast }\left(1\right)=-168+168\left(1\right)-36(1{)}^2=-36\end{array}$

Therefore, the toy car is the farthest at  $t=2$