A bead moves along straight horizonal wire of length `L`, starting from the left end with a velocity `v_(0)`. Its retardation is proportional to the distance from the right end of the wire. Find the initial retardation (in `m//s^(2)`) (at let end of the wire) if the bead reaches the right end of the wire with a velocity `(v_(0))/2`. (given `v_(0)=5m//s` and `L=1m`)

Explanation

To solve the problem, we need to find the initial retardation of the bead as it moves along the wire. The retardation is proportional to the distance from the right end of the wire. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The bead starts from the left end of a wire of length \( L \) with an initial velocity \( v_0 \).
- The bead reaches the right end of the wire with a final velocity of \( \frac{v_0}{2} \).
- The retardation \( a \) is proportional to the distance from the right end, which can be expressed as:
\[ a = -k (L - x) \]
where \( k \) is a constant of proportionality, \( L \) is the length of the wire, and \( x \) is the distance traveled by the bead from the left end.

2. **Using the Work-Energy Principle**:
- The work done by the retardation force is equal to the change in kinetic energy of the bead.
- The initial kinetic energy \( KE_i \) when the bead starts is:
\[ KE_i = \frac{1}{2} m v_0^2 \]
- The final kinetic energy \( KE_f \) when the bead reaches the right end is:
\[ KE_f = \frac{1}{2} m \left(\frac{v_0}{2}\right)^2 = \frac{1}{2} m \frac{v_0^2}{4} = \frac{1}{8} m v_0^2 \]
- The change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_f - KE_i = \frac{1}{8} m v_0^2 - \frac{1}{2} m v_0^2 = \left(\frac{1}{8} - \frac{4}{8}\right) m v_0^2 = -\frac{3}{8} m v_0^2 \]

3. **Calculating the Work Done by Retardation**:
- The work done by the retardation force over the distance \( L \) is given by:
\[ W = \int_0^L (-k(L - x)) \, dx \]
- Evaluating the integral:
\[ W = -k \int_0^L (L - x) \, dx = -k \left[ Lx - \frac{x^2}{2} \right]_0^L = -k \left[ L^2 - \frac{L^2}{2} \right] = -k \frac{L^2}{2} \]

4. **Setting Work Equal to Change in Kinetic Energy**:
- From the work-energy principle, we have:
\[ -k \frac{L^2}{2} = -\frac{3}{8} m v_0^2 \]
- Rearranging gives:
\[ k = \frac{3 m v_0^2}{4 L^2} \]

5. **Finding Initial Retardation**:
- The initial retardation \( a_0 \) at \( x = 0 \) (left end of the wire) is:
\[ a_0 = -k L \]
- Substituting for \( k \):
\[ a_0 = -\left(\frac{3 m v_0^2}{4 L^2}\right) L = -\frac{3 m v_0^2}{4 L} \]

6. **Substituting Given Values**:
- Given \( v_0 = 5 \, \text{m/s} \) and \( L = 1 \, \text{m} \):
\[ a_0 = -\frac{3 m (5)^2}{4 \cdot 1} = -\frac{3 m \cdot 25}{4} = -\frac{75 m}{4} \, \text{m/s}^2 \]

### Final Answer:
The initial retardation at the left end of the wire is:
\[ \frac{75}{4} \, \text{m/s}^2 \]