Question

The tires of a car make 70 revolutions as the car reduces its speed uniformly from 94.0 km/h to 62.0 km/h. The tires have a diameter of 0.86 m. If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Answer #1

Vi = 94 km/hr = 26.11 m/sec

Vf = 62 km/hr = 17.22 m/sec

distance traveled = 2*pi*r*theta = 2*pi*0.43*70 = 189.12 m

Acceleration will be

Vf^2 = Vi^2 + 2*a*d

a = (Vf^2 - Vi^2)/2*d

a = (17.22^2 - 26.11^2)/(2*189.12) = -1.02 m/sec^2

time taken to travel this distance will be

t = (Vf - Vi)/a

t = (17.22 - 26.11)/(-1.02) = 8.71 sec

Time required to stop by same de-acceleration

t1 = (0 - 17.22)/(-1.02) = 16.88 sec

Total time taken = 8.71 + 16.88 = 25.59 sec

Total distance traveled will be

D = 0.5*Vi*T = 0.5*26.11*25.59

**D = 334.07 m**

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