Chapter 10.1, Problem 46E

To determine

To calculate: The average rate of change over the interval $E=0$ to $E=6$ .

Answer to Problem 46E

The required average of changes of Object A and Object B are 0.28 meter per second and 0.33 meter per second.

Explanation of Solution

Given information:

The graph of velocity for object B:

  

The velocity of an object: $v\left(E\right)=\sqrt{\frac{2E}{m}}$

The mass of Object A = 4 kilograms

The given velocity of an object is $v\left(E\right)=\sqrt{\frac{2E}{m}}$ .

For Object A:

Substitute 0 for E , 4 for m in the above equation.

  $\begin{array}{c}v\left(0\right)=\sqrt{\frac{2\left(0\right)}{4}}\\ =\sqrt{\frac{0}{4}}\\ =0\end{array}$

Substitute 6 for E and 4 for m in the above equation.

  $\begin{array}{c}v\left(6\right)=\sqrt{\frac{2\left(6\right)}{4}}\\ =\sqrt{\frac{12}{4}}\\ =\sqrt{3}\\ \approx 1.7\end{array}$

The average rate of change is,

  $\begin{array}{c}\frac{S\left(6\right)-S\left(0\right)}{6-0}=\frac{1.7-0}{6}\\ =\frac{1.7}{6}\\ \approx 0.28\end{array}$

For Object B:

The average rate of change is,

  $\begin{array}{c}\frac{S\left(6\right)-S\left(0\right)}{6-0}=\frac{2-0}{6}\\ =\frac{2}{6}\\ \approx 0.33\end{array}$

Hence, the required average of changes of Object A and Object B are 0.28 meter per second and 0.33 meter per second.