A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. It kinetic energy K changes with time as `dK//dt = gamma t,` where `gamma` is a positive constant of appropriate dimensions. Which of the following statement is (are) true?

Options

A. The force applied on the particle is constant
B. The speed of the particle is proportional to time
C. The distance of the particle from the origin increases linearly with time
D. The force is conservative

Explanation

To solve the problem, we need to analyze the relationship between kinetic energy, velocity, acceleration, and the distance traveled by the particle under the influence of a force. Let's break it down step by step.

### Step 1: Understanding the Change in Kinetic Energy
We are given that the rate of change of kinetic energy \( K \) with respect to time \( t \) is given by:
\[ \frac{dK}{dt} = \gamma t \]
where \( \gamma \) is a positive constant.

### Step 2: Finding Kinetic Energy as a Function of Time
To find the kinetic energy \( K \) as a function of time, we integrate the expression:
\[ dK = \gamma t \, dt \]
Integrating both sides from \( 0 \) to \( t \):
\[ K(t) - K(0) = \int_0^t \gamma t' \, dt' \]
Since the particle is initially at rest, \( K(0) = 0 \):
\[ K(t) = \gamma \int_0^t t' \, dt' = \gamma \left[ \frac{(t')^2}{2} \right]_0^t = \frac{\gamma t^2}{2} \]

### Step 3: Relating Kinetic Energy to Velocity
The kinetic energy \( K \) is also related to the mass \( m \) and velocity \( v \) of the particle by the formula:
\[ K = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = \frac{\gamma t^2}{2} \]
Multiplying both sides by 2:
\[ mv^2 = \gamma t^2 \]
Dividing by \( m \):
\[ v^2 = \frac{\gamma}{m} t^2 \]
Taking the square root:
\[ v = \sqrt{\frac{\gamma}{m}} t \]

### Step 4: Finding Acceleration
To find the acceleration \( a \), we differentiate the velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( \sqrt{\frac{\gamma}{m}} t \right) = \sqrt{\frac{\gamma}{m}} \]
This shows that the acceleration is constant.

### Step 5: Finding the Force
Using Newton's second law, the force \( F \) acting on the particle is given by:
\[ F = ma = m \cdot \sqrt{\frac{\gamma}{m}} = \sqrt{\gamma m} \]
This indicates that the force is constant.

### Step 6: Finding Distance Traveled
To find the distance \( x \) traveled by the particle, we use the relationship:
\[ v = \frac{dx}{dt} \]
Substituting for \( v \):
\[ \frac{dx}{dt} = \sqrt{\frac{\gamma}{m}} t \]
Integrating both sides:
\[ dx = \sqrt{\frac{\gamma}{m}} t \, dt \]
Integrating from \( 0 \) to \( x \) and \( 0 \) to \( t \):
\[ x = \sqrt{\frac{\gamma}{m}} \int_0^t t' \, dt' = \sqrt{\frac{\gamma}{m}} \left[ \frac{(t')^2}{2} \right]_0^t = \frac{\sqrt{\frac{\gamma}{m}}}{2} t^2 \]
This shows that \( x \) is proportional to \( t^2 \), not linearly dependent on \( t \).

### Conclusion
Based on our analysis:
1. The speed of the particle is proportional to time \( t \) (True).
2. The distance of the particle from the origin increases linearly with time (False, it increases with \( t^2 \)).
3. The force acting on the particle is constant (True).
4. The force is conservative (True, as it does not depend on the path).

### Summary of True Statements:
- The speed of the particle is proportional to time.
- The force acting on the particle is constant.
- The force is conservative.