If a constant force acts on a body initially at rest, the distance moved by the body in time t is proportional to

Options

A. t
B. `t^2`
C. `t^3`
D. `t^4`

Explanation

To solve the problem, we need to determine how the distance moved by a body under a constant force, starting from rest, relates to time. We can use the principles of Newton's second law and kinematics to derive the answer step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a body initially at rest, which means its initial velocity (u) is 0.
- A constant force (F) is acting on the body.

2. **Applying Newton's Second Law**:
- According to Newton's second law, the acceleration (a) of the body is given by:
\[ a = \frac{F}{m} \]
where \( F \) is the constant force applied, and \( m \) is the mass of the body. Since both \( F \) and \( m \) are constants, the acceleration \( a \) is also constant.

3. **Using the Kinematic Equation**:
- The kinematic equation for displacement (s) when starting from rest is:
\[ s = ut + \frac{1}{2} a t^2 \]
- Since the body is initially at rest, \( u = 0 \). Therefore, the equation simplifies to:
\[ s = 0 \cdot t + \frac{1}{2} a t^2 \]
- This further simplifies to:
\[ s = \frac{1}{2} a t^2 \]

4. **Identifying Proportionality**:
- From the equation \( s = \frac{1}{2} a t^2 \), we see that \( s \) is proportional to \( t^2 \). This means:
\[ s \propto t^2 \]

5. **Conclusion**:
- Therefore, the distance moved by the body in time \( t \) is proportional to \( t^2 \).

### Final Answer:
The distance moved by the body in time \( t \) is proportional to \( t^2 \).

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