Chapter 12.2, Problem 44E

To determine

Find the velocity when t=2 seconds, using given position function.

Answer to Problem 44E

Velocity at $t=2$ second is $-64$ ft/sec.

Explanation of Solution

Given:

The position function is:

  $s(t)=-16t^2+256$

Definition of velocity at $t=a$ seconds

  $\underset{t\to 0}{\lim }\frac{s(a)-s(t)}{a-t}$

Consider the following position function

  $s(t)=-16t^2+256$

To evaluate the velocity at $t=1$ , proceed as follows.

Use the definition of velocity at $t=a$ seconds.

Definition of velocity at $t=a$ seconds

  $\underset{t\to 0}{\lim }\frac{s(a)-s(t)}{a-t}$

Find the value of $s(2)-s(t)$

  $s(t)=-16t^2+\mathrm{256........}(1)$

Put $s=1$

  $\begin{array}{l}s(1)=-16{(}^2+256\\ =-64+256\\ s(1)=\mathrm{192.........}(2)\end{array}$

Subtract equation (1) from equation (2),

  $\begin{array}{l}s(1)-s(t)=192-\left[-16t^2+256\right]\\ =192+16t^2-256\\ =16t^2-64\\ =16\left(t^2-4\right)\end{array}$

Now divide the expression by $2-t$

  $\begin{array}{l}\frac{s(1)-s(t)}{2-t}=\frac{16\left(t^2-4\right)}{2-t}\\ =\frac{-16(t+2)(2-t)}{(2-t)}\\ =-16(2+t)\end{array}$

Now substitute t=2

  $\begin{array}{cc}\underset{x\to 1}{\lim }\frac{s(2)-s(t)}{2-t}& =\underset{\to \to 1}{\lim }-16(2+t)\\ & \begin{array}{l}=-16(2+2)\\ =-64\end{array}\\ & \end{array}$

Hence, velocity at $t=2$ second is $-64$ ft/sec.