Question

The brakes in a car increase in temperature by ΔT when bringing the car to rest from a speed v. How much greater would ΔT be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.

a) ΔT would be half.

b) ΔT would be the same.

c) ΔT would be double.

d) ΔT would be quadruple.

Answer 1

The temperature increase of a car's brakes, represented by ΔT, would be quadruple if the car's initial speed were doubled because the kinetic energy, which converts into heat energy during braking, is proportional to the square of the velocity.

Step-by-step explanation:

The temperature increase in a car's brakes, denoted by ΔT, when the car comes to a stop from a speed v, is directly related to the kinetic energy that is converted into heat energy. The kinetic energy of an object moving at speed v is given by the formula KE = ½mv², where m is the mass of the object and v is its velocity. When the car initially has twice the speed, its kinetic energy would be KE = ½m(2v)² = 2²½mv² = 4(½mv²), which is four times greater. Therefore, assuming no heat is lost to the surroundings, the increase in temperature would also be four times greater.

Hence, the correct answer to how much greater ΔT would be if the car initially had twice the speed is quadruple.