A model rocket rests on a frictionless horizontal surface and is joined by a string of length `l` to a fixed point so that the rocket moves in a horizontal circular path of radius `l`. The string will break if its tensiion exceeds a value `T`. The rocket engine provides a thrust `F` of constant magnitude along the rocket's direction of motion. the rocket has a mass `m` that does not change with time. Answer the following questions based on the above passage.
What distance does the rocket travel between the time `t_(1)` when the string breaks and the time `2t_(1)`? The rocket engine continues of operate after the string breaks.

Options

A. `(3lT)/(2F)`
B. `(2lT)/(3F)`
C. `(lT)/(2F)`
D. `(2lT)/F`

Explanation

To solve the problem, we need to determine the distance the rocket travels between the time \( t_1 \) (when the string breaks) and \( t_2 \) (which is \( 2t_1 \)).

### Step-by-Step Solution:

1. **Understanding the Situation**:
- The rocket is moving in a circular path of radius \( l \) and experiences a thrust \( F \) from its engine.
- The string will break when the tension exceeds \( T \).

2. **Finding the Time \( t_1 \)**:
- The tension in the string can be expressed in terms of the centripetal force required to keep the rocket moving in a circular path and the tangential force due to the thrust.
- At the moment the string breaks, the centripetal force is given by \( \frac{mv^2}{l} \) and the tangential force is \( F \).
- The condition for the string breaking can be set up as:
\[ T = \frac{mv^2}{l} + F \]
- Rearranging gives us:
\[ \frac{mv^2}{l} = T - F \]
- Thus, we can express \( v \) (the velocity of the rocket at \( t_1 \)):
\[ v = \sqrt{\frac{(T - F)l}{m}} \]

3. **Calculating the Velocity at \( t_1 \)**:
- The thrust \( F \) provides a tangential acceleration \( a_t = \frac{F}{m} \).
- The velocity at time \( t_1 \) can also be expressed as:
\[ v = a_t \cdot t_1 = \frac{F}{m} \cdot t_1 \]

4. **Equating the Two Expressions for Velocity**:
- We can set the two expressions for \( v \) equal to each other:
\[ \sqrt{\frac{(T - F)l}{m}} = \frac{F}{m} \cdot t_1 \]
- Squaring both sides gives:
\[ \frac{(T - F)l}{m} = \frac{F^2}{m^2} t_1^2 \]
- Rearranging for \( t_1^2 \):
\[ t_1^2 = \frac{(T - F)l m}{F^2} \]

5. **Finding the Distance Traveled Between \( t_1 \) and \( t_2 \)**:
- The time interval between \( t_1 \) and \( t_2 \) is \( t_1 \).
- The initial velocity \( u \) at \( t_1 \) is \( v = \frac{F}{m} t_1 \).
- The distance \( s \) traveled during time \( t_1 \) can be calculated using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
- Substituting \( u \), \( t \), and \( a \):
\[ s = \left(\frac{F}{m} t_1\right) t_1 + \frac{1}{2} \left(\frac{F}{m}\right) t_1^2 \]
- This simplifies to:
\[ s = \frac{F}{m} t_1^2 + \frac{1}{2} \frac{F}{m} t_1^2 = \frac{3F}{2m} t_1^2 \]

6. **Substituting \( t_1^2 \)**:
- Now, substituting the expression for \( t_1^2 \):
\[ s = \frac{3F}{2m} \cdot \frac{(T - F)l m}{F^2} \]
- The \( m \) cancels out:
\[ s = \frac{3(T - F)l}{2F} \]

### Final Answer:
The distance traveled by the rocket between the time \( t_1 \) when the string breaks and the time \( t_2 \) is:
\[ s = \frac{3(T - F)l}{2F} \]