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Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g = 10 m/s2)ย
(a)ย ย ย 30m
(b)ย ย ย 40m
ย ย ย ย ย ย ย (c)ย ย ย 12m
(d)ย ย ย 20m
(b)Here u = 72 km/h = 20 m/s; v = 0; a =- ?g = -0.5 × 10 =-5m/s2 As v2 = u1 + 2as, ? s= ((v^2- u^2 ))/2a =((0- (20)^2)/(2× (-5) )=40 m
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