A constant power `P` is applied to a particle of mass `m`. The distance traveled by the particle when its velocity increases from `v_(1)` to `v_(2)` is (neglect friction):

Options

A. `(3P)/m (v_2^2 - v_1^2)`
B. `(m)/(3P) (v_2 - v_1)`
C. `m/(3P) (v_2^3 - v_1^3)`
D. `m/(3P) (v_2^2 - v_1^2)`

Explanation

To solve the problem, we need to find the distance traveled by a particle when its velocity increases from \( v_1 \) to \( v_2 \) under a constant power \( P \). Here’s a step-by-step solution:

### Step 1: Understand the relationship between power, force, and velocity
Power \( P \) is defined as the product of force \( F \) and velocity \( v \):
\[ P = F \cdot v \]
Since force \( F \) can also be expressed as \( F = m \cdot a \) (where \( a \) is acceleration), we can rewrite the power equation as:
\[ P = m \cdot a \cdot v \]

### Step 2: Express acceleration in terms of velocity and distance
Acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can express \( a \) in terms of velocity and distance:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]
Thus, we can rewrite the power equation as:
\[ P = m \cdot \left(\frac{dv}{dx} \cdot v\right) \cdot v = m \cdot v \cdot \frac{dv}{dx} \cdot v = m \cdot v^2 \cdot \frac{dv}{dx} \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ \frac{P}{m} = v^2 \cdot \frac{dv}{dx} \]
This can be further rearranged to separate variables:
\[ \frac{dx}{dv} = \frac{m \cdot v^2}{P} \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side will integrate with respect to \( x \) from \( 0 \) to \( s \), and the right side will integrate with respect to \( v \) from \( v_1 \) to \( v_2 \):
\[ \int_0^s dx = \int_{v_1}^{v_2} \frac{m \cdot v^2}{P} dv \]
This simplifies to:
\[ s = \frac{m}{P} \int_{v_1}^{v_2} v^2 dv \]

### Step 5: Evaluate the integral
The integral of \( v^2 \) is:
\[ \int v^2 dv = \frac{v^3}{3} \]
Applying the limits from \( v_1 \) to \( v_2 \):
\[ s = \frac{m}{P} \left[ \frac{v^3}{3} \right]_{v_1}^{v_2} = \frac{m}{P} \left( \frac{v_2^3}{3} - \frac{v_1^3}{3} \right) \]
This simplifies to:
\[ s = \frac{m}{3P} (v_2^3 - v_1^3) \]

### Final Result
Thus, the distance traveled by the particle when its velocity increases from \( v_1 \) to \( v_2 \) is:
\[ s = \frac{m}{3P} (v_2^3 - v_1^3) \]

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