Chapter 3.1, Problem 66PS

Solution

The speed of the river to transport a particle that is twice as massive as usual, 10 times as massive and 100 times as massive.

  $1.12$ meter per second

  $1.47$ meter per second

  $2.15$ meter per second

Given:

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain river normally flows at a speed of 1 meter per second.

Concept Used:

• The even inter root of a number is defined for positive values only that are greater than or equal to 0.
• The odd integer root of all real numbers is defined.
• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• The equivalent expression in rational exponent for any nth radical is given by, $\sqrt[n]{a}=a^{\frac{1}{n}}$ .

Calculation:

Let the mass of the usual particle be m and the normal speed of the river be s .

Since the mass of the particles that a river can transport is proportional to the sixth power of the speed of the river, it follows

  $\begin{array}{l}m\propto s^6\\ m=k{s}^6\end{array}$

Where k is some real proportionality constant.

Since the river normally flows at a speed of 1 meter per second, it follows

  $\begin{array}{c}m=k{\left(1\right)}^6\\ \Rightarrow m=k\end{array}$

In order to find the speed of the river when the particle is twice as massive as usual, we have

  $\begin{array}{c}m=k{s}^6\\ 2m=m{s}^6\\ s^6=2\\ s=\sqrt[6]{2}\\ s\approx 1.12\end{array}$

So, the speed of the river should be about $1.12$ meter per second.

Similarly, for a particle that is 10 times as massive as usual, it follows

  $\begin{array}{c}m=k{s}^6\\ 10m=m{s}^6\\ s^6=10\\ s=\sqrt[6]{10}\\ s\approx 1.47\end{array}$

So, the speed of the river should be about $1.47$ meter per second.

And, for a particle that is 100 times as massive as usual, it follows

  $\begin{array}{c}m=k{s}^6\\ 100m=m{s}^6\\ s^6=100\\ s=\sqrt[6]{100}\\ s\approx 2.15\end{array}$

Thus, the speed of the river should be about $2.15$ meter per second.