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A walker's speed, v, is proportional to the ratio of his leg length, L, and the period of the...

Question:

A walker's speed v is proportional to the ratio of his leg length L and the period of the repeating motion of his legs T that is, v is proportional to L/T. If the period is measured to be proportional to Lp, where p = 3/5, what power of L must the speed be proportional to?

Proportional relationships:

Two variables are said to be proportional to each other when the change in one affects the other. For example, when we say y is directly proportional to x,

{eq}y \ \propto \ x {/eq}

this means y increases with increasing x, or y decreases with decreasing x. When 'y is inversely proportional to x,

{eq}y \ \propto \ \dfrac{1}{x} {/eq}

this means y increases with decreasing x, or y decreases with increasing x.

Answer and Explanation:

In our case, we have that v is proportional to the ratio of L to T,

{eq}v \ \propto \ \dfrac{L}{T} {/eq}

and that T is proportional to L raised to the power p,

{eq}T \ \propto \ L^p {/eq}

where {eq}p \ = \ \dfrac{3}{5} {/eq}. This means that

{eq}T \ \propto \ L^{\dfrac{3}{5}} {/eq}

and substituting this into the first proportionality expression, we get

{eq}\begin{align*} v \ &\propto \ \dfrac{L}{T}\\ \\ v \ &\propto \ \dfrac{L}{L^\left ({\dfrac{3}{5}}\right )}\\ \\ v \ &\propto \ \left ( L \ \times \ L^\left ({\dfrac{- \ 3}{5}}\right ) \right )\\ \\ v \ &\propto \ L^\left ({1 \ - \ \dfrac{3}{5}}\right )\\ \\ v \ &\propto \ L^\left ({\dfrac{2}{5}}\right ) \end{align*} {/eq}

Therefore, the speed must be proportional to L raised to the power {eq}\dfrac{2}{5} {/eq}.


Learn more about this topic:

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Linear & Direct Relationships

from Physics: High School

Chapter 1 / Lesson 8
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