Question

A car initially traveling at 25.7 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 0.350 m?


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

Answer 1

a. Initial speed Vi = 25.7 m/s

Final speed Vf =0

Acceleration a = - 1.50 m/s^2

Let S is distance traveled by the before stop after the break is applied.

( Vf )^2 = (Vi)^2 + 2aS

or, 0 = (25.7)^2 - (2×1.5×S)

or, S = (25.7)^2/ 3

so, S = 220.2 m

Radii of tyres R = 0.35 m

NO OF REVOLUTIONS

N = S/2R

= 220.2/ 2× 0.35  

= 100.13

b. Let V is linear speed at half distance

V^2 = (Vi)^2 +2a(S/2)

or, V^2 = ( 25.7)^2 - 1.5 × 220.2

or, V^2 = 330.2

Or, V = (330.2)^0.5 m/s

= 18.2 m/s

So, angular speed is

= V/R

= 18.2/0.35 rad/s

= 52.0 rad/S