The driver of a bus travelling with a speed 'v' suddenly observes a wall in front of his bus at a distance 'a'. Father of this driver, who was also a driver, had advised him to take circular turn to avoid hitting in such a situation. However, the driver in question decides otherwise by using his own wisdom. He applies brakes as hard as possible without taking a circular turn, then :

Options

A. he is more likely to hit the wall
B. he is less likely to hit the wall
C. he is as likey to hit the wall
D. none of the above

Answer

To solve the problem, we need to analyze the situation where a bus driver applies brakes to avoid hitting a wall. We will derive the necessary equations and compare the effects of applying brakes versus taking a circular turn.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The bus is traveling with an initial speed \( v \).
- The distance to the wall is \( a \).
- The driver decides to apply brakes instead of taking a circular turn.

2. **Frictional Force and Deceleration**:
- The frictional force \( f \) that helps in stopping the bus is given by:
\[ f = \mu \cdot mg \]
where \( \mu \) is the coefficient of friction, \( m \) is the mass of the bus, and \( g \) is the acceleration due to gravity.
- The deceleration \( a_c \) due to friction can be expressed as:
\[ a_c = -\mu g \]

3. **Using the Kinematic Equation**:
- We can use the kinematic equation to find out if the bus can stop before hitting the wall:
\[ v^2 = u^2 + 2as \]
Here, \( v \) is the final velocity (0 when the bus stops), \( u \) is the initial velocity (\( v \)), \( a \) is the deceleration (\(-\mu g\)), and \( s \) is the stopping distance (\( a \)).
- Substituting the known values:
\[ 0 = v^2 + 2(-\mu g)(a) \]
Rearranging gives:
\[ v^2 = 2\mu g a \]

4. **Finding the Coefficient of Friction**:
- From the equation, we can express the coefficient of friction \( \mu \):
\[ \mu = \frac{v^2}{2ga} \]

5. **Considering the Circular Turn**:
- If the driver had taken a circular turn, the centripetal acceleration \( a_c \) required to make the turn is given by:
\[ a_c = \frac{v^2}{R} \]
where \( R \) is the radius of the circular turn (which is equal to \( a \) in this case).
- The frictional force required for the turn can be expressed as:
\[ \mu' = \frac{v^2}{ga} \]

6. **Comparing the Coefficients of Friction**:
- We now compare the coefficients of friction:
\[ \mu < \mu' \]
- This implies that the friction required to stop the bus is less than that required to take a circular turn.

7. **Conclusion**:
- Therefore, applying the brakes is a safer option as it requires less friction compared to taking a circular turn, which would require a higher coefficient of friction to avoid skidding.

### Summary:
The driver made the right decision by applying the brakes instead of taking a circular turn, as the coefficient of friction required to stop the bus is less than that required to take a turn.