A car initially traveling at 28.9 m/s undergoes a constant negative acceleration

Question

A car initially traveling at 28.9 m/s undergoes a constant negative acceleration of magnitude 1.50 m/s2 after its brakes are applied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?
     rev                      

(b) What is the angular speed of the wheels when the car has traveled half the total distance?
     rad/s

(a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?
     rev                      

(b) What is the angular speed of the wheels when the car has traveled half the total distance?
     rad/s

Explanation / Answer

a) v^2=0
v^2= u^2 - 2as
u^2/ 2a= s
28.9^2/(2 x 1.50)

s= 278.403m

now on tire

0.330 x 2 = diameter = 0.660m
0.660 x pi = 2.07345m circumference
278.403/2.07345 = 134.27 revs

134 rev (3 sig. figure)

b)
Half the distance = 139.201 m

v^2 = u^2 + 2as
(28.9^2) + ( 2 x (-1.5) x 139.201)

v^2= 417.607

v= 20.4354 m/s
20.4354/2.07345 = 9.85 rps
9.85 x 2pi = 61.8 rad/sec

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