When a `2` kg car driven at `20 m//s` on a level road is suddenly put into netural gear (i.e. allowed to coast), the velocity decreases in the following manner:
`V=20/(1+(t/20))m//s`
Where t is time in sec. The deceleration of car at the instant its speed is `10 m//s` is.

Options

A. `1//4 m//s^(2)`
B. `1//2 m//s^(2)`
C. `1 m//^(2)`
D. `3//4 m//s^(2)`

Explanation

To solve the problem, we need to find the deceleration of a car that is coasting and has its velocity described by the equation \( V = \frac{20}{1 + \frac{t}{20}} \) m/s. We are specifically interested in the deceleration at the moment when the car's speed is \( 10 \) m/s.

### Step-by-Step Solution:

1. **Set the velocity equation equal to 10 m/s**:
\[ V = \frac{20}{1 + \frac{t}{20}} = 10 \]

2. **Cross-multiply to solve for \( t \)**:
\[ 10(1 + \frac{t}{20}) = 20 \]
\[ 10 + \frac{10t}{20} = 20 \]
\[ \frac{10t}{20} = 20 - 10 \]
\[ \frac{10t}{20} = 10 \]
\[ t = 20 \text{ seconds} \]

3. **Differentiate the velocity function to find acceleration**:
The acceleration \( a \) is given by the derivative of velocity with respect to time:
\[ a = \frac{dV}{dt} \]
We differentiate \( V = \frac{20}{1 + \frac{t}{20}} \):
\[ a = \frac{d}{dt} \left( \frac{20}{1 + \frac{t}{20}} \right) \]
Using the quotient rule:
\[ a = \frac{0 \cdot (1 + \frac{t}{20}) - 20 \cdot \frac{1}{20} \cdot 1}{(1 + \frac{t}{20})^2} \]
Simplifying gives:
\[ a = -\frac{400}{(20 + t)^2} \]

4. **Substitute \( t = 20 \) seconds into the acceleration equation**:
\[ a = -\frac{400}{(20 + 20)^2} = -\frac{400}{40^2} = -\frac{400}{1600} = -\frac{1}{4} \text{ m/s}^2 \]

5. **Determine the deceleration**:
Since deceleration is the negative of acceleration when speed is decreasing:
\[ \text{Deceleration} = -a = -(-\frac{1}{4}) = \frac{1}{4} \text{ m/s}^2 \]

### Final Answer:
The deceleration of the car at the instant its speed is \( 10 \) m/s is \( \frac{1}{4} \text{ m/s}^2 \).

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