Question

Race-track turns are often "banked" (tilted inward) so that cars can take them at high speed without skidding. Consider a circular track 2km in length banked at an angle of 20o , and just for fun suppose the track is covered in ice (after a bad storm, lets say). With what speed does a car have to drive in order to make it around the track?

Answer 1

To calculate the speed a car needs to drive around a banked curve, we can use the equation for centripetal force. Rearranging the equation, we can solve for the speed using the radius of the curve and the angle of the banked curve.

Step-by-step explanation:

In order to calculate the speed at which a car needs to drive to make it around a banked curve, we can use the equation for centripetal force. The centripetal force is equal to the force of gravity component perpendicular to the road surface. This can be represented by the equation:

Fc = m * v^2 / r

where Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the curve. Rearranging the equation, we can solve for v:

v = sqrt(r * g * tan(theta))

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and theta is the angle of the banked curve.

Using this equation, we can calculate the speed at which the car needs to drive to make it around the track.

Answer 2

Answer:33.7 m/s

Step-by-step explanation:

Given

circumference of track= 2km=2000 m

thus radius r=318.26 m

Banking
`=20^(\circ)`

Let R be the reaction force

Thus
`R\sin \theta =(mv^2)/(r)`---1

`R\cos theta =mg`----2

divide 1 & 2 we get

`\tan \theta =(v^2)/(rg)`

`v=√(rg\tan \theta )`

v=33.7 m/s