A car is moving with a constant accelaration of `4_(m//s^(2))`. On seeing a person infront of it the driver suddenly applies a brake . IF the car is moving with a velolcity of `20_(m//s)` at the moment when driver sees the person and the reaction time of the driver is 2s, then the distance it travels after the movement it sees the person and the just before applying the break is

Options

A. 48m
B. 40m
C. 8M
D. `45m

Explanation

To solve the problem step by step, we will calculate the distance traveled by the car during the driver's reaction time before the brakes are applied.

### Step 1: Understand the Given Information
- Initial velocity of the car (u) = 20 m/s
- Acceleration of the car (a) = 4 m/s² (though this won't affect the distance during the reaction time)
- Reaction time (t) = 2 seconds

### Step 2: Calculate the Distance Traveled During Reaction Time
During the reaction time, the car continues to move at its initial velocity since the brakes have not yet been applied. The formula for distance traveled when there is no acceleration is:

\[ s = u \cdot t \]

Where:
- \( s \) = distance traveled
- \( u \) = initial velocity
- \( t \) = time

### Step 3: Substitute the Values
Now, we can substitute the values into the formula:

\[ s = 20 \, \text{m/s} \times 2 \, \text{s} \]

### Step 4: Perform the Calculation
Calculating the above expression:

\[ s = 20 \times 2 = 40 \, \text{m} \]

### Step 5: Conclusion
The distance traveled by the car during the reaction time before the brakes are applied is **40 meters**.

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