Divide a plane `10 m` long and `5 m` high into three parts so that a body starting from rest takes equal times to slide down these. Also find the time taken then.

Explanation

To solve the problem of dividing a plane that is 10 m long and 5 m high into three parts such that a body starting from rest takes equal times to slide down each part, we will follow these steps:

### Step 1: Determine the acceleration of the body
The acceleration of the body sliding down the inclined plane is given by the component of gravitational acceleration acting along the incline.

1. The height (h) of the plane is 5 m, and the length (L) of the plane is 10 m.
2. The angle of inclination (θ) can be found using the sine function:
\[ \sin(\theta) = \frac{\text{height}}{\text{length}} = \frac{5}{10} = 0.5 \]
Thus, \( \theta = 30^\circ \).
3. The acceleration (a) down the incline is:
\[ a = g \sin(\theta) = 10 \times 0.5 = 5 \, \text{m/s}^2 \]

### Step 2: Apply the equations of motion
We will use the equation of motion \( s = ut + \frac{1}{2} a t^2 \) for each segment of the incline.

1. For the first segment (x1), starting from rest (u = 0):
\[ x_1 = 0 \cdot t + \frac{1}{2} \cdot 5 \cdot t^2 = \frac{5}{2} t^2 \]

2. For the first two segments combined (x1 + x2), the time taken is \( 2t \):
\[ x_1 + x_2 = 0 \cdot 2t + \frac{1}{2} \cdot 5 \cdot (2t)^2 = \frac{1}{2} \cdot 5 \cdot 4t^2 = 10t^2 \]

3. For all three segments combined (x1 + x2 + x3), the time taken is \( 3t \):
\[ x_1 + x_2 + x_3 = 0 \cdot 3t + \frac{1}{2} \cdot 5 \cdot (3t)^2 = \frac{1}{2} \cdot 5 \cdot 9t^2 = \frac{45}{2} t^2 \]

### Step 3: Set up the equations
Now we have three equations:
1. \( x_1 = \frac{5}{2} t^2 \)
2. \( x_1 + x_2 = 10t^2 \)
3. \( x_1 + x_2 + x_3 = 10 \)

### Step 4: Solve for time (t)
From the third equation:
\[ 10 = \frac{45}{2} t^2 \]
Solving for t:
\[ t^2 = \frac{20}{45} = \frac{4}{9} \]
Thus,
\[ t = \frac{2}{3} \, \text{seconds} \]

### Step 5: Calculate distances x1, x2, and x3
1. Substitute \( t^2 = \frac{4}{9} \) into the first equation:
\[ x_1 = \frac{5}{2} \cdot \frac{4}{9} = \frac{10}{9} \, \text{m} \]

2. Substitute \( t^2 \) into the second equation to find \( x_2 \):
\[ x_1 + x_2 = 10t^2 \implies \frac{10}{9} + x_2 = 10 \cdot \frac{4}{9} = \frac{40}{9} \]
Thus,
\[ x_2 = \frac{40}{9} - \frac{10}{9} = \frac{30}{9} \, \text{m} \]

3. Finally, calculate \( x_3 \):
\[ x_3 = 10 - (x_1 + x_2) = 10 - \left( \frac{10}{9} + \frac{30}{9} \right) = 10 - \frac{40}{9} = \frac{50}{9} \, \text{m} \]

### Final Results
- \( x_1 = \frac{10}{9} \, \text{m} \)
- \( x_2 = \frac{30}{9} \, \text{m} \)
- \( x_3 = \frac{50}{9} \, \text{m} \)
- Time taken \( t = \frac{2}{3} \, \text{seconds} \)