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…

Question

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=10m/s^2)

Accepted Answer

Correct answer: B. (b) Step 1: Calculation of acceleration due to friction force For stopping the car in shortest distance without skidding(sliding), the maximum static friction should act in opposite direction. So , ,f= μN ⇒f= μmg ( From figure ,N= mg ) Now, Applying Newton's second law on car along x-direction (Rightwards Positive) F_x= ma_x ⇒-f= ma ⇒- μmg= ma ⇒a=- μg=-0.5 10=-5 ms^-2 Negative means left direction (retardation) Step 2: Calculation of Stopping distance Since Retardation is constant, so applying kinematics equation of motion: v^2=u^2+2as Puta=-5 ms^-2 and initial velocityu=72 5/18=20 m/s ⇒o=2O^2-2 5 s ∴s=40 m Shortest distance in which car can be stopped is 40 m

$Consider a car moving along a straight horizontal road with a speed of 72 km/h If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g=10m/ s^{2})$

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