The distance travelled by a motor car in `t` seconds 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. `10.08ft`
B. `10ft`
C. `11ft`
D. `11.5ft`

Explanation

To find the distance travelled by the motor car before it stops, we start with the given equation for distance:

\[ s = 22t - 12t^2 \]

### Step 1: Find the velocity
The velocity \( v \) of the car is given by the derivative of the distance \( s \) with respect to time \( t \):

\[ v = \frac{ds}{dt} = \frac{d}{dt}(22t - 12t^2) \]

Calculating the derivative:

\[ v = 22 - 24t \]

### Step 2: Set the velocity to zero to find the time when the car stops
To find when the car stops, we set the velocity \( v \) to zero:

\[ 22 - 24t = 0 \]

Solving for \( t \):

\[ 24t = 22 \]

\[ t = \frac{22}{24} = \frac{11}{12} \text{ seconds} \]

### Step 3: Substitute \( t \) back into the distance equation
Now, we substitute \( t = \frac{11}{12} \) back into the distance equation to find the total distance travelled before the car stops:

\[ s = 22\left(\frac{11}{12}\right) - 12\left(\frac{11}{12}\right)^2 \]

Calculating each term:

1. First term:
\[ 22 \times \frac{11}{12} = \frac{242}{12} \]

2. Second term:
\[ 12 \times \left(\frac{11}{12}\right)^2 = 12 \times \frac{121}{144} = \frac{1452}{144} = \frac{121}{12} \]

### Step 4: Combine the results
Now we combine the two terms:

\[ s = \frac{242}{12} - \frac{121}{12} = \frac{121}{12} \text{ feet} \]

Thus, the distance travelled by the car before it stops is:

\[ \boxed{\frac{121}{12} \text{ feet}} \]