To solve the problem step by step, we will follow the reasoning laid out in the video transcript and apply the relevant physics concepts.
### Step 1: Understand the Given Information
- Tangential acceleration, \( a_c = 0.62 \, \text{m/s}^2 \)
- Radius of the circular path, \( R = 40 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.2 \)
- Initial velocity, \( v_0 = 0 \)
### Step 2: Use the Kinematic Equation
Since the car starts from rest and moves with constant tangential acceleration, we can use the kinematic equation:
\[
v^2 = v_0^2 + 2a_c s
\]
Here, \( v_0 = 0 \), so the equation simplifies to:
\[
v^2 = 2a_c s \tag{1}
\]
### Step 3: Determine the Maximum Velocity Before Sliding
The maximum velocity before sliding occurs can be determined using the frictional force, which provides the necessary centripetal force. The centripetal acceleration \( a_c \) is given by:
\[
a_c = \frac{v^2}{R}
\]
The frictional force can be expressed as:
\[
F_r = \mu mg
\]
Setting the centripetal force equal to the frictional force gives:
\[
\frac{v^2}{R} = \mu g
\]
Rearranging this, we find:
\[
v^2 = \mu g R \tag{2}
\]
### Step 4: Substitute Values
Substituting the known values into equation (2):
- \( g \approx 9.81 \, \text{m/s}^2 \)
- \( R = 40 \, \text{m} \)
- \( \mu = 0.2 \)
Calculating \( v^2 \):
\[
v^2 = 0.2 \times 9.81 \times 40
\]
\[
v^2 = 0.2 \times 392.4 = 78.48 \, \text{m}^2/\text{s}^2
\]
### Step 5: Substitute \( v^2 \) Back into Equation (1)
Now, substitute \( v^2 \) from equation (2) into equation (1):
\[
78.48 = 2 \times 0.62 \times s
\]
Solving for \( s \):
\[
s = \frac{78.48}{2 \times 0.62} = \frac{78.48}{1.24} \approx 63.23 \, \text{m}
\]
### Step 6: Relate \( s \) to \( x \)
According to the problem, the distance \( s \) is equal to \( 12x \):
\[
12x = 63.23
\]
Solving for \( x \):
\[
x = \frac{63.23}{12} \approx 5.27 \, \text{m}
\]
### Final Answer
Thus, the value of \( x \) is approximately:
\[
\boxed{5.27 \, \text{m}}
\]
