A body falls from a certain bhight. Two seconds later another body falls from the same height. How long after the beginning of motion of the first body is the distance between the bodies twice the distance at the moment the second body starts to fall ?

Options

A. 3 s
B. 10 s
C. 15 s
D. 20 s

Answer

To solve the problem step by step, we will analyze the motion of both bodies and use the equations of motion.

### Step 1: Understand the scenario
- Let the first body fall from a height \( h \) at time \( t = 0 \).
- The second body starts falling 2 seconds later, at \( t = 2 \) seconds.

### Step 2: Determine the distance fallen by the first body
Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
For the first body, the initial velocity \( u = 0 \):
\[ s_1 = 0 \cdot t + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]
So, the distance fallen by the first body after \( t \) seconds is:
\[ s_1 = \frac{1}{2} g t^2 \]

### Step 3: Determine the distance fallen by the second body
The second body starts falling at \( t = 2 \) seconds, so at time \( t \), the time for the second body falling is \( t - 2 \):
\[ s_2 = 0 \cdot (t - 2) + \frac{1}{2} g (t - 2)^2 = \frac{1}{2} g (t - 2)^2 \]

### Step 4: Find the distance between the two bodies
The distance between the two bodies at time \( t \) is:
\[ d = s_1 - s_2 = \frac{1}{2} g t^2 - \frac{1}{2} g (t - 2)^2 \]

### Step 5: Simplify the expression for distance
Expanding \( s_2 \):
\[ s_2 = \frac{1}{2} g (t^2 - 4t + 4) = \frac{1}{2} g t^2 - 2gt + 2g \]
Now substituting back into the distance equation:
\[ d = \frac{1}{2} g t^2 - \left(\frac{1}{2} g t^2 - 2gt + 2g\right) \]
\[ d = 2gt - 2g \]

### Step 6: Set up the equation for the condition given in the problem
According to the problem, we want the distance \( d \) to be twice the distance fallen by the second body at the moment it starts falling (which is \( s_2 \) when \( t = 2 \)):
\[ s_2 = \frac{1}{2} g (2 - 2)^2 = 0 \]
Thus, we need \( d = 2 \cdot 0 = 0 \) at the moment the second body starts falling. However, we need to find when \( d = 2 \cdot s_2 \) at some later time.

### Step 7: Set the equation for the distance condition
We need:
\[ d = 2 \cdot \left(\frac{1}{2} g (2 - 2)^2\right) \]
This simplifies to:
\[ d = 2 \cdot 0 = 0 \]
This is not what we want. We need to find the time when the distance \( d \) is twice the distance fallen by the second body after it starts falling.

### Step 8: Set up the final equation
We want:
\[ d = 2 \cdot s_2 \]
Substituting:
\[ 2gt - 2g = 2 \cdot \frac{1}{2} g (t - 2)^2 \]
\[ 2gt - 2g = g(t^2 - 4t + 4) \]
Dividing through by \( g \) (assuming \( g \neq 0 \)):
\[ 2t - 2 = t^2 - 4t + 4 \]
Rearranging gives:
\[ t^2 - 6t + 6 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ t = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3} \]

### Step 10: Conclusion
We take the positive root since time cannot be negative:
\[ t = 3 + \sqrt{3} \text{ seconds} \]