A curve in a road 8m wide has a radius of 60 m. How much higher than its inner edge should be the other edge of the road be if it is to be bunked properly at 20 km/hr?
change km/hr to m/s
Net force from centripetal and gravity shouuld be a some angle theta.
Tan theta= v^2/r / g= v^2/rg
solve for theta
Now, the outer edge:
SinTheta=outeredgeincrease/8m
solve for outer edge increase
Well, if the road wants to be bunked properly at 20 km/hr, I think it's gonna need a good night's sleep and maybe a cup of coffee in the morning. As for the height of the other edge, let's just say it should be high enough to make the road feel like it's on top of the world! But seriously, since we're dealing with a curved road, the height will depend on the angle of the curve and the velocity of the vehicle. Without those specifics, it's hard to give you an exact answer.
To determine how much higher the outer edge of the road should be than its inner edge, we need to calculate the banking angle.
The formula for the banking angle (θ) is given by:
θ = arctan(v^2 / (g * r))
where:
v = velocity of the vehicle (converted to m/s),
g = acceleration due to gravity, which is approximately 9.8 m/s^2,
r = radius of the curve.
First, let's convert the velocity from km/hr to m/s:
20 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 5.56 m/s (rounded to two decimal places).
Now, substitute the values in the formula:
θ = arctan((5.56)^2 / (9.8 * 60))
θ ≈ arctan(0.5507) (rounded to four decimal places)
Use a calculator or trigonometric table to find the arctan value:
θ ≈ 29.94° (rounded to two decimal places)
This means that the road should be banked at an angle of approximately 29.94°.
To determine how much higher the outer edge of the road should be than the inner edge, we need to consider the banking angle required for the road curve. In this case, the speed at which the road is to be banked is given as 20 km/hr.
To calculate the banking angle, we can use the following formula:
tan(θ) = v^2 / (g * r)
Where:
- θ is the banking angle in radians
- v is the speed in meters per second (m/s)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- r is the radius of the curve in meters
First, we need to convert the speed from km/hr to m/s. Since 1 km = 1000 meters and 1 hour = 3600 seconds, we can calculate:
v = 20 km/hr * (1000 m/1 km) * (1 hr/3600 s) = (20 * 1000) / 3600 ≈ 5.56 m/s
Now we can substitute the values into the formula:
tan(θ) = (5.56 m/s)^2 / (9.8 m/s^2 * 60 m)
tan(θ) ≈ 0.03
To find the banking angle θ, we need to take the arctan of both sides:
θ ≈ arctan(0.03)
Using a scientific calculator, we find θ ≈ 1.72 degrees.
To determine the difference in height between the inner and outer edges of the road, we can use trigonometry. Since the road is 8 meters wide, the height difference between the two edges can be calculated as follows:
Height difference = (8 m / 2) * tan(θ)
Height difference = 4 m * tan(1.72 degrees)
Using the tangent function, we find Height difference ≈ 0.12 m or approximately 12 cm.
Therefore, the outer edge of the road should be approximately 12 cm higher than the inner edge to be properly banked at a speed of 20 km/hr.