Eli went for his driving test in an ice storm. He told the registry tester not to worry. He can negotiate an exit ramp at 35 km/hr that has a radius of curvature of 95m. a) What was his speed in meter/second? b)Find required angle for his banked curve to handle an ice filled road without the car slipping?
a) 9.7 m/s ?
b)?
draw the figure.
up the ramp, the centriptetal force is v^2/r * cosTheta
down the ramp gravity=mg*sinTheta
mg*sintheta=mv^2/r * cosTheta
tantheta= v^2/rg
check that.
thank you
To find the required angle for the banked curve, we can use the equation for the angle of banking on a curved road:
tan(θ) = (v^2)/(g * r)
where:
- θ is the angle of banking
- v is the velocity of the car
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- r is the radius of curvature of the curve
Let's calculate the speed of Eli's car first.
a) To find the speed in meters per second (m/s), we need to convert the speed from kilometers per hour (km/hr) to meters per second.
Speed in m/s = (Speed in km/hr) * (1000 meters / 1 kilometer) * (1 hour / 3600 seconds)
Given that Eli's speed is 35 km/hr, we can calculate:
Speed in m/s = 35 km/hr * (1000 m / 1 km) * (1 hr / 3600 s)
Speed in m/s = 35 * 1000 / 3600
Speed in m/s ≈ 9.7 m/s
So, Eli's speed is approximately 9.7 m/s.
Now, let's calculate the required angle for the banked curve.
b) Substituting the values into the equation, we get:
tan(θ) = (v^2) / (g * r)
tan(θ) = (9.7^2) / (9.8 * 95)
Using a calculator, we can find:
tan(θ) ≈ 0.1005
To find the angle, we can take the inverse tangent (arctan or tan^-1) of both sides:
θ ≈ arctan(0.1005)
Using a calculator, we find:
θ ≈ 5.738 degrees
Therefore, the required angle for the banked curve to handle the ice-filled road without the car slipping is approximately 5.738 degrees.