A rat is running on ice with speed `v=pims^(-1)`. Suddenly he decides to turn by `90^(@)` and want of keep running with the same speed throughout. What is the least amount of time (in seocnds) he needs for such a turn? Suppose that rat's feet can move independently. Coefficieny of friction between rat's feet and ice is 0.125.
`("Given : "pi^(2)=g)`

Answer

To solve the problem of the rat turning 90 degrees while maintaining its speed on ice, we can follow these steps:

### Step 1: Understand the Forces Acting on the Rat
When the rat is running and decides to turn, it experiences a centripetal force that allows it to change direction. The friction between the rat's feet and the ice provides this centripetal force.

### Step 2: Set Up the Equation for Centripetal Force
The centripetal force required for circular motion is given by:
\[ F_c = \frac{mv^2}{R} \]
where:
- \( m \) is the mass of the rat,
- \( v \) is the speed of the rat,
- \( R \) is the radius of the turn.

The frictional force that provides this centripetal force is:
\[ F_f = \mu mg \]
where:
- \( \mu \) is the coefficient of friction (0.125),
- \( g \) is the acceleration due to gravity.

### Step 3: Equate the Forces
Setting the centripetal force equal to the frictional force gives us:
\[ \frac{mv^2}{R} = \mu mg \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{R} = \mu g \]

### Step 4: Solve for the Radius \( R \)
Rearranging the equation to find \( R \):
\[ R = \frac{v^2}{\mu g} \]
Given that \( g = \pi^2 \) and \( \mu = 0.125 \), we can substitute these values in:
\[ R = \frac{v^2}{0.125 \cdot \pi^2} \]

### Step 5: Calculate the Distance for the Turn
The distance \( d \) the rat travels while turning 90 degrees (a quarter of a circle) is:
\[ d = \frac{1}{4} \cdot 2\pi R = \frac{\pi R}{2} \]

### Step 6: Substitute \( R \) into the Distance Formula
Substituting \( R \) from the previous step:
\[ d = \frac{\pi}{2} \cdot \frac{v^2}{0.125 \cdot \pi^2} = \frac{v^2}{0.25 \pi} \]

### Step 7: Use the Speed to Find Time
Using the relationship \( v = \frac{d}{t} \), we can solve for time \( t \):
\[ t = \frac{d}{v} = \frac{v^2}{0.25 \pi v} = \frac{v}{0.25 \pi} = \frac{4v}{\pi} \]

### Step 8: Substitute the Given Speed
Given \( v = 5 \, \text{m/s} \):
\[ t = \frac{4 \cdot 5}{\pi} = \frac{20}{\pi} \]

### Step 9: Calculate the Numerical Value
Using \( \pi \approx 3.14 \):
\[ t \approx \frac{20}{3.14} \approx 6.37 \, \text{seconds} \]

### Final Answer
The least amount of time the rat needs to turn 90 degrees while maintaining its speed is approximately:
\[ \boxed{6.37 \, \text{seconds}} \]