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 was the magnitude of instantaneous net acceleration at time `t_(1)//2`? Obtain answer in terms of `F, T` and `m`.

Options

A. `([T^(2)+8F^(2)]^(1//2))/m`
B. `([T+4F^(2)]^(1//2))/(2m)`
C. `([T^(2)+16F^(2)]^(1//2))/(4m)`
D. none of these

Explanation

To solve the problem, we need to find the magnitude of the instantaneous net acceleration of the rocket at time \( t = \frac{t_1}{2} \). We will consider both the centripetal acceleration and the tangential acceleration due to the thrust provided by the rocket engine.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Rocket:**
- The thrust \( F \) acts in the direction of the rocket's motion.
- The tension \( T \) in the string provides the necessary centripetal force to keep the rocket moving in a circular path.

2. **Determine the Velocity of the Rocket at Time \( t = \frac{t_1}{2} \):**
- The initial velocity \( u = 0 \) (the rocket starts from rest).
- The acceleration \( a \) can be expressed as \( a = \frac{F}{m} \).
- Using the equation of motion \( v = u + at \), we find:
\[ v_1 = 0 + \left(\frac{F}{m}\right) \left(\frac{t_1}{2}\right) = \frac{F t_1}{2m} \]

3. **Calculate the Centripetal Acceleration \( a_c \):**
- The centripetal acceleration is given by:
\[ a_c = \frac{v^2}{r} \]
- Substituting \( v_1 \) and \( r = l \):
\[ a_c = \frac{\left(\frac{F t_1}{2m}\right)^2}{l} = \frac{F^2 t_1^2}{4m^2 l} \]

4. **Relate Tension to Centripetal Force:**
- The tension \( T \) provides the centripetal force necessary for circular motion:
\[ T = m a_c = m \left(\frac{F^2 t_1^2}{4m^2 l}\right) = \frac{F^2 t_1^2}{4m l} \]

5. **Determine the Tangential Acceleration \( a_t \):**
- The tangential acceleration is simply given by:
\[ a_t = \frac{F}{m} \]

6. **Calculate the Net Instantaneous Acceleration \( A \):**
- The net acceleration is the vector sum of the centripetal and tangential accelerations:
\[ A = \sqrt{a_c^2 + a_t^2} \]
- Substituting the values we found:
\[ A = \sqrt{\left(\frac{F^2 t_1^2}{4m^2 l}\right)^2 + \left(\frac{F}{m}\right)^2} \]
- This simplifies to:
\[ A = \sqrt{\frac{F^4 t_1^4}{16 m^4 l^2} + \frac{F^2}{m^2}} = \sqrt{\frac{F^2}{m^2} \left(\frac{F^2 t_1^4}{16 l^2} + 1\right)} \]
- Therefore, the final expression for the instantaneous net acceleration is:
\[ A = \frac{F}{m} \sqrt{\frac{F^2 t_1^4}{16 l^2} + 1} \]

### Final Answer:
The magnitude of the instantaneous net acceleration at time \( t = \frac{t_1}{2} \) is:
\[ A = \frac{F}{m} \sqrt{\frac{F^2 t_1^4}{16 l^2} + 1} \]