A motor car moving with a uniform speed of `20 m//sec` comes to stop on the application of brakes after travelling a distance of 10m Its acceleration is

Options

A. `20m//sec^(2)`
B. `-20m//sec^(2)`
C. `-40m//sec^(2)`
D. `+2m//sec^(2)`

Answer

To find the acceleration of the motor car that comes to a stop after traveling a distance of 10 meters, we can use the kinematic equation:

\[ V^2 = U^2 + 2as \]

Where:
- \( V \) = final velocity (0 m/s, since the car comes to a stop)
- \( U \) = initial velocity (20 m/s)
- \( a \) = acceleration (which we need to find)
- \( s \) = distance traveled (10 m)

### Step-by-step Solution:

1. **Identify the known values:**
- Initial velocity, \( U = 20 \, \text{m/s} \)
- Final velocity, \( V = 0 \, \text{m/s} \)
- Distance traveled, \( s = 10 \, \text{m} \)

2. **Substitute the known values into the kinematic equation:**
\[ 0^2 = (20)^2 + 2a(10) \]

3. **Simplify the equation:**
\[ 0 = 400 + 20a \]

4. **Rearrange the equation to solve for acceleration \( a \):**
\[ 20a = -400 \]

5. **Divide both sides by 20:**
\[ a = \frac{-400}{20} = -20 \, \text{m/s}^2 \]

6. **Conclusion:**
The acceleration of the motor car is \( -20 \, \text{m/s}^2 \).