A bit of mud stuck to a bicycle's front wheel of radius r detaches and is flung horizontally forward when it is at the top of the wheel. The bicycle is moving forward at a speed v and it is rolling without slipping. The horizontal distance travelled by the mud after detaching from the wheel is:

Explanation

To solve the problem of the horizontal distance traveled by the mud after detaching from the bicycle's front wheel, we can follow these steps:

### Step 1: Understand the motion of the mud
When the mud detaches from the top of the wheel, it has an initial horizontal velocity equal to the linear velocity of the top of the wheel. Since the bicycle is rolling without slipping, the velocity of the top of the wheel is twice the velocity of the center of the wheel. Therefore, the initial horizontal velocity \( u \) of the mud is:
\[ u = 2v \]

### Step 2: Determine the height from which the mud falls
The mud detaches from the top of the wheel, which is at a height equal to the diameter of the wheel (since it is at the top). The height \( h \) from which the mud falls is:
\[ h = 2r \]

### Step 3: Calculate the time of flight
To find the time \( t \) it takes for the mud to fall to the ground, we can use the second equation of motion for vertical motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = 2r \), \( u = 0 \) (initial vertical velocity), and \( a = g \) (acceleration due to gravity). Thus, the equation simplifies to:
\[ 2r = 0 + \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{4r}{g} \]
Taking the square root, we find:
\[ t = \sqrt{\frac{4r}{g}} = \frac{2\sqrt{r}}{\sqrt{g}} \]

### Step 4: Calculate the horizontal distance traveled
The horizontal distance \( d \) traveled by the mud can be calculated using the formula:
\[ d = u \cdot t \]
Substituting the values we have:
\[ d = (2v) \cdot \left(\frac{2\sqrt{r}}{\sqrt{g}}\right) = \frac{4v\sqrt{r}}{\sqrt{g}} \]

### Final Result
The horizontal distance traveled by the mud after detaching from the wheel is:
\[ d = \frac{4v\sqrt{r}}{\sqrt{g}} \]