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. (a)The force applied on the particle is constant
B. (b)The speed of the particle is proportional to time
C. (c)The distance of the particle from the origin increases linearly with time
D. (d)The force is conservative

Explanation

To solve the problem step by step, we will analyze the given information and derive the necessary relationships.

### Step 1: Understand the given information
We have a particle of mass \( m \) that is initially at rest at the origin. The change in kinetic energy \( K \) with respect to time \( t \) is given by:
\[ \frac{dK}{dt} = \gamma t \]
where \( \gamma \) is a positive constant.

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

### Step 3: Relate kinetic energy to velocity
The kinetic energy \( K \) is also related to the velocity \( v \) of the particle:
\[ K = \frac{1}{2} mv^2 \]
Equating the two expressions for kinetic energy:
\[ \frac{1}{2} mv^2 = \frac{\gamma t^2}{2} \]
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: Analyze the velocity and acceleration
From the expression for velocity:
\[ v = \sqrt{\frac{\gamma}{m}} t \]
we can see that velocity is directly proportional to time \( t \).

### Step 5: Find acceleration
Acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( \sqrt{\frac{\gamma}{m}} t \right) = \sqrt{\frac{\gamma}{m}} \]
Since \( a \) is constant, the force \( F \) acting on the particle can be calculated using Newton's second law:
\[ F = ma = m \cdot \sqrt{\frac{\gamma}{m}} = \sqrt{\gamma m} \]
This indicates that the force is constant.

### Step 6: Determine the displacement
To find the displacement \( x \), we can integrate the velocity:
\[ dx = v \, dt = \sqrt{\frac{\gamma}{m}} t \, dt \]
Integrating from \( 0 \) to \( t \):
\[ x(t) - x(0) = \int_0^t \sqrt{\frac{\gamma}{m}} t' \, dt' = \sqrt{\frac{\gamma}{m}} \left[ \frac{(t')^2}{2} \right]_0^t = \sqrt{\frac{\gamma}{m}} \cdot \frac{t^2}{2} \]
Thus:
\[ x(t) = \frac{\sqrt{\gamma}}{2\sqrt{m}} t^2 \]

### Conclusion
Now we can summarize the findings:
1. The force applied on the particle is constant.
2. The speed of the particle is proportional to time.
3. The distance from the origin increases with the square of time, not linearly.

### Final Statements
- The correct statements are:
- The force applied on the particle is constant.
- The speed of the particle is proportional to time.
- The distance of the particle from the origin does not increase linearly with time (it increases with \( t^2 \)).
- The force is conservative.