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A car starts from rest. It has to cover a distance of 500 m. the coefficient of friction between the road and the tyre is 0.5. The minimum time in which the car can cover this distance is (g=10m/s 2 )
As soon as the car sets into motion the friction force acting on its tyre would be equal to the limiting value of the static friction which is equal to 5m (where m is the mass of the car).So this force provides the acc which is equal to 5m/s^2..so the minimum time required is about 14 sec.
friction = u(mg) ma = umg a = ug =5m/s2 (retardation) now we have V2 = U2 - 2aS S=500 , V=0 , U=? ,a =5 after substituting these we get U = (5000)1/2 now we have , V = U -at V =0 , U = (5000)1/2 , a=5 substituting these we get t = (200)1/2 = 10sqrt2 sec
friction = u(mg)
ma = umg
a = ug =5m/s2 (retardation)
now we have
V2 = U2 - 2aS
S=500 , V=0 , U=? ,a =5
after substituting these we get
U = (5000)1/2
now we have ,
V = U -at
V =0 , U = (5000)1/2 , a=5
substituting these we get
t = (200)1/2 = 10sqrt2 sec
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