A motorbike driver applies brakes when the bike is moving at 95 km/hr and reduce

Question

A motorbike driver applies brakes when the bike is moving at 95 km/hr and reduces its velocity uniformly until it stops in 5.6 seconds. If the front tire makes 61 revolutions during this time what is its diameter?

I know the answer is 0.39 m but I am not sure how to get there. Please include any equations and how you derived those equations you use and explain where the numbers are coming from, be as detailed as you possibly can be. thanks

AmritPalGoyal please do not answer this question again because you are leaving out information and not responding to my questions in the comments, thank you

Explanation / Answer

u = 95 km/hr = 26.3889 m/s

t = 5.6 s

v =0,

a = u/t = 4.71 m/s^2,

s= u^2/2a = 73.889 m

1 revolution = pi* D..(horizontal distance)

so,

73.8889 = 61* pi* D ,

D = 0.3857 m

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