A helicopter takes off along the vertical with an acceleration `a = 3m//s^(2)` and zero initial velocity. In a certain time the pilot switches off the engine. At the point of take off, the sound dies away in a time `t_(2) = 30 sec`. Determine the velocity of the helicopter at the moment when its engine is switched off assuming that velocity of sound is 320 m/s.

Answer

To solve the problem, we need to determine the velocity of the helicopter at the moment when its engine is switched off. We will follow these steps:

### Step 1: Determine the distance traveled by the helicopter before the engine is switched off.

The helicopter takes off with an initial velocity \( u = 0 \) and an acceleration \( a = 3 \, \text{m/s}^2 \). The distance \( s \) traveled by the helicopter in time \( t \) can be calculated using the equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the values:

\[ s = 0 \cdot t + \frac{1}{2} \cdot 3 \cdot t^2 = \frac{3}{2} t^2 \]

### Step 2: Relate the time taken by the sound to travel the same distance.

When the engine is switched off, the sound takes time \( t' \) to travel the distance \( s \) back to the ground. The speed of sound is given as \( v_s = 320 \, \text{m/s} \). The distance \( s \) can also be expressed in terms of the time \( t' \):

\[ s = v_s \cdot t' = 320 \cdot t' \]

### Step 3: Establish the relationship between \( t \) and \( t' \).

According to the problem, the total time from when the helicopter takes off to when the sound reaches the ground is given as:

\[ t + t' = 30 \, \text{seconds} \]

From this, we can express \( t' \) in terms of \( t \):

\[ t' = 30 - t \]

### Step 4: Substitute \( t' \) into the distance equation.

Now, we can substitute \( t' \) into the equation for distance \( s \):

\[ \frac{3}{2} t^2 = 320 \cdot (30 - t) \]

### Step 5: Rearranging the equation.

Expanding and rearranging gives:

\[ \frac{3}{2} t^2 + 320t - 9600 = 0 \]

To eliminate the fraction, we can multiply the entire equation by 2:

\[ 3t^2 + 640t - 19200 = 0 \]

### Step 6: Solve the quadratic equation.

We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = 640 \), and \( c = -19200 \).

Calculating the discriminant:

\[ b^2 - 4ac = 640^2 - 4 \cdot 3 \cdot (-19200) \]
\[ = 409600 + 230400 = 640000 \]

Now, applying the quadratic formula:

\[ t = \frac{-640 \pm \sqrt{640000}}{2 \cdot 3} \]
\[ = \frac{-640 \pm 800}{6} \]

Calculating the two possible values for \( t \):

1. \( t = \frac{160}{6} = \frac{80}{3} \, \text{seconds} \) (valid)
2. \( t = \frac{-1440}{6} = -240 \, \text{seconds} \) (not valid)

### Step 7: Calculate the velocity of the helicopter.

Now that we have \( t = \frac{80}{3} \), we can find the velocity \( v \) of the helicopter at the moment the engine is switched off using the formula:

\[ v = u + at \]

Substituting the known values:

\[ v = 0 + 3 \cdot \frac{80}{3} = 80 \, \text{m/s} \]

### Final Answer:

The velocity of the helicopter at the moment when its engine is switched off is \( 80 \, \text{m/s} \).

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