Question

If the co-efficient of friction between a car tyres and a roadway is 0.62, what is the least distance in which the car can accelerate from rest to 20.7 m/s ?

Answer 1

we know that

coefficient of friction =force of friction/ Normal force

therefore

0.62 = m×a/m× g

a = 0.62 ×9.8= 6.07 m/s^2

and

we know that

$s = {v}^{2} - u ^{2} \div 2a \\ ={20.7}^{2} - {0}^{2} \div 2 \times 6.07 \\ = 35.1 \: m$