Chapter 12.8, Problem 12OE

To determine

To Describe: Translate the statement into formula:

The acceleration, $a$ , of a moving object is directly proportional to the distance , $d$ , it travels and inversely proportional to the square of time, $t$ , it travels.

Answer to Problem 12OE

The formula of the statement is $a=k\frac{d^2}{t^2}$ .

Explanation of Solution

Given information:

Translate statement to formula:

The acceleration, $a$ , of a moving object is directly proportional to the distance , $d$ , it travels and inversely proportional to the square of time, $t$ , it travels.

Use $K$ as constant.

Consider the statement,

The acceleration, $a$ , of a moving object is directly proportional to the distance , $d$ , it travels and inversely proportional to the square of time, $t$ , it travels.

Here if a variable varies directly as he product of two or more other variables the resulting relationship is a joint variation.

As , $a$ ,varies directly to the distance $d$ and inversely to the square of time , $t$ .

  $a$ is directly to the distance, $d$ .

  $a\alpha d$ .

Then $a$ is inversely as its square of time, $t$ .

  $a\alpha \frac{1}{t^2}$

From both of the equations:

  $\begin{array}{c}a\alpha d\\ a\alpha \frac{1}{t^2}\\ a\alpha \frac{d}{t^2}\end{array}$

Same as above,

Here,

  $a\alpha \frac{d}{t^2}$

For replacing $\alpha$ apply constant $K$ .

  $\begin{array}{l}a\alpha \frac{d}{t^2}\\ a=K\frac{d}{t^2}\end{array}$

Thus, the formula is: $a=k\frac{d^2}{t^2}$ .