A car with constant tangential acceleration At=0.80m/s2 along a horizontol surface circumscribing a circle of radius R = 40m. The co-efficient of friction between the wheels of the car and the surface is u = 0.20. What distance will the car move without sliding if its initial velocty is zero ?

To determine the distance a car will move without sliding when it has a constant tangential acceleration, we need to consider the forces acting on the car, particularly the frictional force that prevents it from sliding. Let's break this down step by step.

Understanding the Forces at Play

When a car accelerates along a circular path, two main forces are involved: the tangential force due to acceleration and the centripetal force required to keep the car moving in a circle. The frictional force between the tires and the road provides both of these forces.

Frictional Force Calculation

The maximum frictional force that can act on the car is given by:

• F_friction = u * N

Where:

• u is the coefficient of friction (0.20 in this case).
• N is the normal force, which equals the weight of the car (mg) on a horizontal surface.

Assuming the mass of the car is m, the normal force is:

• N = mg

Thus, the maximum frictional force becomes:

• F_friction = u * mg = 0.20 * mg

Centripetal Force Requirement

The centripetal force required to keep the car moving in a circle is given by:

• F_centripetal = (mv^2) / R

Where v is the velocity of the car and R is the radius of the circle (40 m). For the car to move without sliding, the frictional force must be equal to or greater than the centripetal force:

• F_friction ≥ F_centripetal
• 0.20 * mg ≥ (mv^2) / R

We can cancel m from both sides (assuming the mass is not zero):

• 0.20g ≥ v^2 / R

Rearranging gives us:

• v^2 ≤ 0.20gR

Finding the Maximum Velocity

Substituting the values for g (approximately 9.81 m/s²) and R (40 m):

• v^2 ≤ 0.20 * 9.81 * 40
• v^2 ≤ 78.48
• v ≤ √78.48 ≈ 8.85 m/s

Distance Calculation with Constant Acceleration

Now that we have the maximum velocity, we can find out how far the car can travel while accelerating at A_t = 0.80 m/s² until it reaches this velocity. Using the kinematic equation:

• v^2 = u^2 + 2a s

Where:

• v is the final velocity (8.85 m/s),
• u is the initial velocity (0 m/s),
• a is the acceleration (0.80 m/s²),
• s is the distance traveled.

Substituting the known values:

• (8.85)^2 = 0 + 2 * 0.80 * s

This simplifies to:

• 78.22 = 1.6s

Solving for s gives:

• s = 78.22 / 1.6 ≈ 48.89 m

Final Thoughts

The car will move approximately 48.89 meters without sliding before it reaches the maximum velocity allowed by the frictional force. This distance ensures that the car maintains traction while navigating the circular path.