A car is moving on a straight horizontal road with a speed of $72 k m p h$ . If the coefficient of kinetic friction between the tyre of the car and the road is $0.5$ , then find the minimum distance, within which the car can be stopped.

72 m

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40 m

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30 m

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20 m

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Solution

The correct option is D $40 m$
From FBD, $m a = f = μ N = μ m g$
or $a = 0.5 (10) = 5 m / s^{2}$
by using $v^{2} - u^{2} = 2 a S$ , we will get,
$0 - (\frac{72 \times 1000}{3600})^{2} = 2 (5) S$
or $400 = 10 S$
$∴$ Distance , $S = 40 m$ .

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A car is moving on a straight horizontal road with a speed of 72kmph. If the coefficient of kinetic friction between the tyre of the car and the road is 0.5, then find the minimum distance, within which the car can be stopped.

The correct option is D 40mFrom FBD, ma=f=μN=μmgor a=0.5(10)=5m/s2by using v2−u2=2aS, we will get,0−(72×10003600)2=2(5)Sor 400=10S∴ Distance ,S=40m.

A car is moving on a straight horizontal road with a speed of $72 k m p h$ . If the coefficient of kinetic friction between the tyre of the car and the road is $0.5$ , then find the minimum distance, within which the car can be stopped.

The correct option is D $40 m$
From FBD, $m a = f = μ N = μ m g$
or $a = 0.5 (10) = 5 m / s^{2}$
by using $v^{2} - u^{2} = 2 a S$ , we will get,
$0 - (\frac{72 \times 1000}{3600})^{2} = 2 (5) S$
or $400 = 10 S$
$∴$ Distance , $S = 40 m$ .