Question

A car initially traveling at 25.7 m/s undergoes a constant negative acceleration of magnitude 1.80 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 0.320 m?
rev

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

Answer 1

Solution

Initial velocity = vi = 25.7 m/s

Deceleration = -1.80 m/s /s = negative acceleration

Radius = r = 0.320m

a ) Angular acceleration of the car = = a/r

=> = --1.8 /0.32 = - 5.625 rad / s / s

Initial angular velocity = i = v/ r = 25.7 / .32 = 80.3 rad/ s

Final angular velocity = 0

f^2 = i^2 + 2

=> = (0^2 - 80.3^2 ) /( 2*-5.625) = 573 .3 rad

No. Of revolutions = 573.3 / 2 pi = 91.3 revolutions

b) Distance travelled = d = ( vf^2 - vi^2 ) / 2 a

= ( 0^2 - 25.7^2) / 2*-1.8

= 183.469 m

Half this distance = 183.469 / 2 = 91.73 m

Vf^2 = (25.7)^2 + (-1.8)(91.73) = 330.245

=> vf = final velocity = 18.2 m/s

Angular speed =vf / r = 18.2 / 0.32 = 56.9 rad/ s