Ahoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline. what is the order in which they arrive at the bottom of the incline?

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

,where h=height of release

a=acceleration

Where I=moment of inertia

a for hoop

a for Uniform solid cylinder

a for spherical shell

a for Uniform Solid

time taken will be inversely proportional to the square root of acceleration

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)