If a certain car, going with speed v₁, rounds a level curve with a radius R₁, it is just on the verge of skidding.If its speed is now doubled, the radius of the tightest curve on the same road that it can round without skidding is: A)2R₁ B)4R₁ C)R₁/2 D)R₁/4 E)R₁

Question

Quizplus · Accepted Answer

The force of friction acting on the car during the turn is given by: friction force = mv^2/R where m is the mass of the car, v is the speed of the car, and R is the radius of the turn. As the car approaches the limit of skidding, the maximum friction force is equal to the force required to keep the car on the track: maximum friction force = m*g*cos(theta) where g is the acceleration due to gravity and theta is the angle between the horizontal and the incline. At the limit of skidding, these two forces are equal: mv^2/R = m*g*cos(theta) Solving for R, we get: R = v^2/(g*cos(theta)) When the speed is doubled, the new speed is 2v. Plugging this in for v, we get: Rnew = (2v)^2/(g*cos(theta)) = 4v^2/(g*cos(theta)) To find the tightest turn the car can make without skidding, we want to maximize the value of cos(theta). This occurs when theta = 0 (i.e. the incline is level). Therefore, we can simplify the expression for Rnew: Rnew = 4v^2/g We can see that Rnew is four times larger than the original radius R, so the answer is choice B) 4R₁.

If a Certain Car, Going with Speed V1, Rounds a Level

If a Certain Car, Going with Speed V₁, Rounds a Level