A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 11.4 s. The diameter of a tire is 66.5 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.

58.6886 revolutions

Explanation:

First we need to know the total distance travelled by the car, and we can do that using Torricelli formula:

V2= Vo2 + 2aDS

V = 21.5

Vo = 0

a = 21.5/11.4 = 1.886

(21.5)^2 = 2*1.886*DS

DS = 462.25/3.772 = 122.5477 m

For each revolution of the tire, the car moves the circunference of the tire, which is pi*d = 3.14*66.5 =  208.81 cm =  2.0881 m

So, to know the number of revolutions, we divide the total travel distance by the circunference of the tire:

122.5477/2.0881 =  58.6886

c) up

hope this

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.