A street car moves rectilinearly from station A to the next station B (from rest to rest) with an acceleration varying according to the law `f = a-bx,` where a and b are constants and x is the distance from station A. The distance between the two stations and the maximum velocity are

Options

A. `x=2a/b,v_(max)=a/(sqrtb)`
B. `x=a/2b,v_(max)=a/b`
C. `x=a/2b,v_(max)=b/(sqrta)`
D. `x=a/b,v_(max)=(sqrta)/b`

Explanation

To solve the problem, we need to find the distance between the two stations (A and B) and the maximum velocity of the streetcar, given that its acceleration varies according to the law \( f = a - bx \), where \( a \) and \( b \) are constants and \( x \) is the distance from station A.

### Step 1: Relate Force to Acceleration
The force \( f \) acting on the streetcar is related to its acceleration \( a \) by Newton's second law:
\[ f = m \cdot a \]
Given that \( f = a - bx \), we can write:
\[ m \cdot a = a - bx \]
From this, we can express acceleration as:
\[ a = \frac{f}{m} = a - \frac{b}{m}x \]

### Step 2: Use the Work-Energy Principle
The work done on the streetcar will be equal to the change in kinetic energy. The work done can be expressed as:
\[ \int_0^x (a - bx) \, dx \]
And the change in kinetic energy from rest to maximum velocity \( v \) is:
\[ \frac{1}{2} mv^2 \]

### Step 3: Integrate the Force
Now we integrate the force over the distance:
\[ \int_0^x (a - bx) \, dx = ax - \frac{b}{2}x^2 \bigg|_0^x = ax - \frac{b}{2}x^2 \]

### Step 4: Set Up the Energy Equation
Setting the work done equal to the change in kinetic energy gives us:
\[ ax - \frac{b}{2}x^2 = \frac{1}{2} mv^2 \]

### Step 5: Solve for Maximum Velocity
At the maximum velocity, we can express \( v^2 \) in terms of \( x \):
\[ v^2 = \frac{2}{m}(ax - \frac{b}{2}x^2) \]

### Step 6: Find the Condition for Maximum Velocity
To find the maximum velocity, we need to find when acceleration \( a - bx = 0 \):
\[ a - bx = 0 \implies x = \frac{a}{b} \]

### Step 7: Substitute Back to Find Maximum Velocity
Substituting \( x = \frac{a}{b} \) into the expression for \( v^2 \):
\[ v^2 = \frac{2}{m}\left(a\frac{a}{b} - \frac{b}{2}\left(\frac{a}{b}\right)^2\right) \]
Simplifying this gives:
\[ v^2 = \frac{2a^2}{mb} - \frac{b}{2}\frac{a^2}{b^2} = \frac{2a^2}{mb} - \frac{a^2}{2b} = \frac{4a^2 - a^2}{2mb} = \frac{3a^2}{2mb} \]
Thus, the maximum velocity is:
\[ v_{max} = \sqrt{\frac{3a^2}{2mb}} = \frac{a\sqrt{3}}{\sqrt{2m}b} \]

### Step 8: Find the Distance Between Stations
Using the condition for the distance between stations, we substitute back into the equation:
\[ x = \frac{2a}{b} \]

### Final Answers
1. The distance between the two stations \( x = \frac{2a}{b} \).
2. The maximum velocity \( v_{max} = \frac{a}{\sqrt{b}} \).