Two banked curves have the same radius. Curve A is banked at an angle of 11°, and curve B is banked at an angle of 16°. A car can travel around curve A without relying on friction at a speed of 18.5 m/s. At what speed can this car travel around curve B without relying on friction?

Well, let's see... If a car can travel around curve A without relying on friction at a speed of 18.5 m/s, then I guess we can call it a "Banked Racer!"😄 Now, for curve B, which is banked at an angle of 16°, we need to figure out the speed.

You see, the angle of banking affects how much the car relies on friction to go around the curve. In this case, curve B is banked at a higher angle than curve A, which means the car will rely on friction more to keep on cruising.

To find the speed, we can use the formula:

v = √(rgtanθ),

where v is the speed, r is the radius of the curve, g is the acceleration due to gravity, and θ is the angle of banking.

Since the radius is the same for both curves, and the acceleration due to gravity is constant, we can simplify the formula to:

v = √(gtanθ).

Plugging in the values, we have:

v = √(9.8 * tan(16°)).

Calculating this, we get:

v ≈ 19.6 m/s.

So, the "Banked Racer" can speed around curve B without relying on friction at approximately 19.6 m/s. Just make sure to buckle up and steel your nerves for the higher banked adventure! 🏎️💨💨

To find the speed at which the car can travel around curve B without relying on friction, we need to use the concept of the critical speed.

The critical speed is the speed at which the car can safely navigate the banked curve without the assistance of friction. It is given by the formula:

V = sqrt(g * R * tanθ)

Where:
V is the critical speed,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
R is the radius of the curve, and
θ is the angle of banking.

Let's plug in the given values for curve B:

g = 9.8 m/s^2,
θ = 16°, and
V = ?

Since both curves have the same radius, we don't need to know the specific value of the radius.

V = sqrt(9.8 * R * tan(16°))

Now we can calculate V.

To find out the speed at which the car can travel around curve B without relying on friction, we can use the concept of centripetal force.

The centripetal force required to keep an object moving in a circular path is given by the equation:

Fc = (mv^2) / r

Where:
- Fc is the centripetal force
- m is the mass of the object
- v is the velocity of the object
- r is the radius of the curved path

Now, let's consider curve A first. Since the car can travel around curve A without relying on friction, the centripetal force required is provided entirely by the banking angle. This means that the force due to the banking angle is equal to the gravitational force acting on the car (mg). Since both curves have the same radius, the radius cancels out in the equation.

For curve A:
Fc = (mv^2) / r = mg (since the car does not rely on friction)
Therefore, (mv^2) / r = mg

Now, let's move on to curve B. We need to consider the force of friction in this case. The force due to the banking angle will only be a part of the required centripetal force. The remaining force will be provided by friction.

For curve B:
Fc = (mv^2) / r = mg sinθ + μmg cosθ
Where:
- μ is the coefficient of friction between the car's tires and the road surface
- θ is the banking angle

Since the car can travel around curve B without relying on friction, we can set the friction term equal to zero:
(mv^2) / r = mg sinθ

We can now set up an equation to find the velocity (v) for curve B without relying on friction:

(mv^2) / r = mg sinθ

And we can solve for v:
v^2 = r g sinθ
v = √(r g sinθ)

Plugging in the given values for curve B:
θ = 16°
r = radius (given as the same for both curves)
g = gravitational acceleration (approximately 9.8 m/s^2)

v = √(r * 9.8 * sin(16°))

After substituting the values and evaluating the expression, you will find the speed at which the car can travel around curve B without relying on friction.

tan α=ma₁/mg = (mv₁²/r)/mg= v₁²/rg

tan β=ma₂/mg = (mv₂²/r)/mg= v₂²/rg

tan α/ tan β= v₁²/v₂²
v₂ = v₁•sqrt(tan β/tan α)=18.5•sqrt(tan16/tan11)=
=22.47 m/s