The velocity of a car moving on a straight road increases linearly accroding to equation, `v=1+bx` , where `a &b` are positive constants. The acceleration in the course of such motion `:(x` is the displacement `)`

Options

A. increases
B. decreases
C. stay constant
D. becomes zero

Explanation

To solve the problem, we need to find the acceleration of the car given the velocity function \( v = 1 + bx \), where \( b \) is a positive constant and \( x \) is the displacement.

### Step-by-Step Solution:

1. **Identify the given equation**:
The velocity of the car is given by the equation:
\[ v = 1 + bx \]

2. **Understand the relationship between acceleration, velocity, and displacement**:
Acceleration \( a \) can be expressed in terms of velocity \( v \) and displacement \( x \) using the formula:
\[ a = v \frac{dv}{dx} \]
Here, \( \frac{dv}{dx} \) is the derivative of velocity with respect to displacement.

3. **Differentiate the velocity function**:
To find \( \frac{dv}{dx} \), we differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = \frac{d}{dx}(1 + bx) = b \]
Since \( b \) is a constant, the derivative is simply \( b \).

4. **Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula**:
Now, we can substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula:
\[ a = v \frac{dv}{dx} = (1 + bx) \cdot b \]

5. **Expand the expression**:
Expanding the expression gives:
\[ a = b + b^2x \]

6. **Analyze the result**:
The expression \( a = b + b^2x \) shows that the acceleration is a function of displacement \( x \). Since \( b \) is a positive constant, \( b^2 \) is also positive. Therefore, as \( x \) increases, the term \( b^2x \) increases, which means that the acceleration \( a \) increases as \( x \) increases.

### Final Answer:
The acceleration of the car increases with displacement.