The distance travelled by a motor car in t second after the brakes are applied is s feet, where s = 22t `-12t^(2)`. The distance travelled by the car before it stops, is

Options

A. 10.08 ft
B. 10 ft
C. 11 ft
D. 11.5 ft

Explanation

To solve the problem, we need to find the distance traveled by a motor car after the brakes are applied, given the equation for distance \( s = 22t - 12t^2 \). The car stops when its velocity becomes zero.

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the total distance traveled by the car before it stops. The distance \( s \) is given as a function of time \( t \).

2. **Differentiate the Distance Function**: To find when the car stops, we need to find the velocity, which is the derivative of the distance with respect to time. Thus, we differentiate \( s \):
\[ \frac{ds}{dt} = \frac{d}{dt}(22t - 12t^2) \]
Using the power rule, we get:
\[ \frac{ds}{dt} = 22 - 24t \]

3. **Set the Velocity to Zero**: The car stops when the velocity is zero:
\[ 22 - 24t = 0 \]

4. **Solve for Time \( t \)**: Rearranging the equation gives:
\[ 24t = 22 \implies t = \frac{22}{24} = \frac{11}{12} \text{ seconds} \]

5. **Substitute \( t \) Back into the Distance Equation**: Now we substitute \( t = \frac{11}{12} \) back into the distance equation to find the total distance traveled:
\[ s = 22\left(\frac{11}{12}\right) - 12\left(\frac{11}{12}\right)^2 \]
Calculating each term:
\[ s = \frac{22 \times 11}{12} - 12 \times \frac{121}{144} \]
Simplifying:
\[ s = \frac{242}{12} - \frac{121}{12} \]
\[ s = \frac{242 - 121}{12} = \frac{121}{12} \]

6. **Convert to Decimal**: To find the numerical value:
\[ s = \frac{121}{12} \approx 10.0833 \text{ feet} \]

Thus, the distance traveled by the car before it stops is approximately **10.08 feet**.