If the velocity of a body moving in a straight line is proportional to the square root of the distance traversed, then it moves with

Options

A. variable force
B. constant force
C. zero force
D. zero acceleration

Explanation

To solve the problem, we need to analyze the relationship between velocity, distance, and acceleration based on the information given. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We are given that the velocity \( v \) of a body is proportional to the square root of the distance \( s \) traversed. This can be mathematically expressed as:
\[ v \propto \sqrt{s} \]
Therefore, we can write:
\[ v = k \sqrt{s} \]
where \( k \) is a constant of proportionality.

**Hint**: Remember that proportionality means we can express one quantity as a constant multiplied by another.

2. **Finding Acceleration**:
Acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
To find \( a \), we need to differentiate \( v \) with respect to \( t \).

3. **Using the Chain Rule**:
We can apply the chain rule for differentiation. Since \( v = k \sqrt{s} \), we differentiate it with respect to \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(k s^{1/2}) = k \cdot \frac{1}{2} s^{-1/2} \cdot \frac{ds}{dt} \]
Here, \( \frac{ds}{dt} \) is the velocity \( v \).

**Hint**: The chain rule allows us to differentiate composite functions. Remember to express \( \frac{ds}{dt} \) in terms of \( v \).

4. **Substituting for Velocity**:
Substituting \( v = k \sqrt{s} \) into the equation:
\[ a = \frac{1}{2} k s^{-1/2} \cdot v \]
Replacing \( v \) with \( k \sqrt{s} \):
\[ a = \frac{1}{2} k s^{-1/2} \cdot (k \sqrt{s}) = \frac{1}{2} k^2 \]
Notice that \( s^{-1/2} \) and \( \sqrt{s} \) cancel out.

**Hint**: When substituting, ensure that you keep track of the variables and their relationships.

5. **Conclusion**:
The expression for acceleration \( a \) simplifies to a constant value \( \frac{1}{2} k^2 \). Since \( a \) is constant, this implies that the force acting on the body, given by \( F = m \cdot a \), is also constant (assuming mass \( m \) is constant).

Therefore, the body moves with **constant force**.

**Hint**: Remember that constant acceleration implies constant force when mass is constant.

### Final Answer:
The body moves with **constant force**.