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 ∝ 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? Enter your answer as a fraction (in lowest terms), first the numerator, then the denominator.
I have no idea how to go about this
It sounds like you mean
period ∝ L^(3/5)
v ∝ L/T = L/L^(3/5) = L^(2/5)
Well, let's break it down step by step. We know that the walker's speed is proportional to the ratio of his leg length, L, and the period of the repeating motion of his legs, T.
We're also given that the period, T, is proportional to L raised to the power of p, where p = 3/5. In other words, T ∝ L^(3/5).
So we can substitute this into the first equation: v ∝ L / T. Since T ∝ L^(3/5), we have v ∝ L / (L^(3/5)).
Now, when we have a fraction in the denominator, we can flip it and make it a negative exponent: v ∝ (L^(5/3)) / L.
Finally, when we divide two powers with the same base, we subtract the exponents: v ∝ L^(5/3 - 1).
Simplifying further, we get v ∝ L^(5/3 - 3/3) = L^(2/3).
So, the power of L that the speed is proportional to is 2/3.
To solve this problem, we need to find out the power of L to which the speed, v, is proportional.
We are given that the period, T, is proportional to Lp, where p = 3/5. This means T ∝ L^(3/5).
We also know that v ∝ L/T. Substituting the value of T, we get v ∝ L/(L^(3/5)).
To simplify this expression, we can rewrite L^(3/5) as (L^(1/5))^3.
So, v ∝ L/(L^(1/5))^3.
Using the property of exponents (a^(m/n) = (n√(a))^m), we can rewrite this as v ∝ L/(L^(1/5))^3 = L/(L^(3/5))^3 = L/(L^(3/5 * 3))^3 = L/L^9/5.
Next, we subtract the exponents when dividing with the same base: L^(a-b) = L^a / L^b.
Thus, v ∝ L/L^(9/5) = L^(1-9/5) = L^(-4/5).
Therefore, the speed is proportional to L^(-4/5). The numerator of the power is -4, and the denominator is 5.
To solve this problem, we need to use the information provided and apply the concept of proportionality. Let's break it down step by step:
1. We are given that the 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. Mathematically, v ∝ L/T.
2. The problem states that the period, T, is proportional to Lp, where p is equal to 3/5.
3. To find the power of L that the speed is proportional to, we need to find the exponent of L in the proportionality of v ∝ L/T.
4. Now, let's substitute the proportionality of T ∝ Lp into our initial equation. We have v ∝ L/(Lp).
5. To simplify the equation, we can rewrite the denominator as (L^(p)) by applying the power rule of exponents. Thus, v ∝ (L^(1-p)).
6. Now, we can see that the power of L in the proportionality of the speed, v, is (1 - p). Therefore, the numerator of the power is 1, and the denominator is p, which is 3/5.
7. In fraction form, the power of L that the speed is proportional to is 1/ (3/5). Simplifying this fraction, we get 5/3 as the answer.
So, the answer is 5/3.