Consider a car moving along a straight horizontal road with a speed of $72 k m / h$ . 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:
(Take $g = 10 m / s^{2}$ )

30 m

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40 m

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72 m

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20 m

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Solution

The correct option is B $40 m$
$v_{i n i t i a l} = u = 72 k m / h = 20 m / s v_{f i n a l} = v = 0 μ = 0.5 a_{d e c e l a r a t i o n} = μ g = 0.5 \times 10 = 5 ㎨ v = 0 u^{2} - 2 a s = 0 400 - 2 \times 5 s = 0 s = 40 m$

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Consider a car moving along a straight horizontal road with a speed of 72km/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:(Take g=10m/s2)

The correct option is B 40mvinitial=u=72km/h=20m/svfinal=v=0μ=0.5adecelaration=μg=0.5×10=5㎨v=0u2−2as=0400−2×5s=0s=40m

Consider a car moving along a straight horizontal road with a speed of $72 k m / h$ . 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:
(Take $g = 10 m / s^{2}$ )

The correct option is B $40 m$
$v_{i n i t i a l} = u = 72 k m / h = 20 m / s v_{f i n a l} = v = 0 μ = 0.5 a_{d e c e l a r a t i o n} = μ g = 0.5 \times 10 = 5 ㎨ v = 0 u^{2} - 2 a s = 0 400 - 2 \times 5 s = 0 s = 40 m$