A cyclest is riding a bicycle on a circular track in a horizontal plane with maximum possible speed on the track. Coefficent of friction between the track and cycle tyre is `mu` on seeing a cat on the track, the cyclist applies brakes so hard that the tyres of the cycle stop rotating abruptly. Let `veca_(1)` and `veca_(2)` be the acceleration of bicycle just befor and just after the brakes applied, then `|bar(a_(2))-bara_(1)|` is

Options

A. `mug`
B. `2mug`
C. zero
D. `sqrt(2)mug`

Answer

To solve the problem, we will analyze the situation step by step.

### Step 1: Understand the scenario
The cyclist is riding on a circular track with maximum speed. The maximum speed is determined by the frictional force between the tires and the track, which provides the necessary centripetal force to keep the bicycle moving in a circle.

### Step 2: Identify the initial acceleration (a1)
Before the brakes are applied, the cyclist experiences centripetal acceleration due to the circular motion. The formula for centripetal acceleration \( a_c \) is given by:

\[ a_c = \frac{v^2}{r} \]

However, in this case, we can also relate it to the frictional force. The maximum frictional force \( F_f \) that provides the centripetal force is given by:

\[ F_f = \mu mg \]

Where:
- \( \mu \) is the coefficient of friction,
- \( m \) is the mass of the bicycle and cyclist,
- \( g \) is the acceleration due to gravity.

Since this frictional force is equal to the centripetal force required for circular motion, we can equate them:

\[ \mu mg = ma_1 \]

Here, \( a_1 \) is the centripetal acceleration. Dividing both sides by \( m \) (mass of the cyclist and bicycle), we get:

\[ a_1 = \mu g \]

### Step 3: Identify the final acceleration (a2)
After the cyclist applies the brakes, the tires stop rotating abruptly. This means that the bicycle is no longer in motion, and the acceleration due to the circular motion becomes zero. Therefore, we have:

\[ a_2 = 0 \]

### Step 4: Calculate the difference in acceleration
Now we need to find the absolute difference between the two accelerations \( |a_2 - a_1| \):

\[ |a_2 - a_1| = |0 - \mu g| = |\mu g| = \mu g \]

### Conclusion
Thus, the difference in acceleration just before and just after the brakes are applied is:

\[ |a_2 - a_1| = \mu g \]

### Final Answer
The answer is \( \mu g \).

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