A car is moving on a straight road with a speed of `20 m//s`. The drive of the car put the breaks at some instant which retards the car uniformly at a rate of `2 m//s^(2)`, until the car stops completely then find the time `(t_(0))` taken by the car and distance `(d)` travelled by car, from the instant the drive put the breaks till it stops:

Options

A. `t_(0) = 10` sec
B. `d = 50 m`
C. `t_(0) = 5` sec
D. `d = 100 m`

Explanation

To solve the problem, we will break it down into steps to find the time taken (t₀) and the distance (d) traveled by the car from the moment the brakes are applied until it stops.

### Step 1: Identify the given values
- Initial velocity (u) = 20 m/s
- Final velocity (v) = 0 m/s (since the car stops)
- Retardation (a) = -2 m/s² (negative because it is a deceleration)

### Step 2: Use the first equation of motion to find time (t₀)
The first equation of motion is:
\[ v = u + at \]

Substituting the known values into the equation:
\[ 0 = 20 + (-2) \cdot t₀ \]

This simplifies to:
\[ 0 = 20 - 2t₀ \]
\[ 2t₀ = 20 \]
\[ t₀ = \frac{20}{2} = 10 \text{ seconds} \]

### Step 3: Use the second equation of motion to find distance (d)
The second equation of motion is:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the known values:
\[ d = 20 \cdot 10 + \frac{1}{2} \cdot (-2) \cdot (10)^2 \]

Calculating each term:
1. \( 20 \cdot 10 = 200 \)
2. \( \frac{1}{2} \cdot (-2) \cdot 100 = -100 \)

Now, combine these results:
\[ d = 200 - 100 = 100 \text{ meters} \]

### Final Results
- Time taken (t₀) = 10 seconds
- Distance traveled (d) = 100 meters