Consider a car moving along a straight horizontal road with a speed of 72 km h- 1 . 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 s- 2 )

Question

Consider a car moving along a straight horizontal road with a speed of 72 km h- 1 . 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 s- 2 ) (A) 30m (B) 40 m (C) 72 m (D) 20 m

Tardigrade Admin · Accepted Answer

Here u=72kmh- 1=20ms- 1;v=0; a=-μ g=-0.5× 10=-5ms- 2 As v2=u2+2 as, ∴ s=((v2 - u2)/2 a)=((. 0 - (20)2 .)/2 × (- 5))=40m

Q. Consider a car moving along a straight horizontal road with a speed of $72 km h^{- 1}$ . 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 $g = 10 m s^{- 2}$ )

$30 m$

$40 m$

$72 m$

$20 m$

Here $u = 72 km h^{- 1} = 20 m s^{- 1}; v = 0;$
$a = - μg = - 0.5 \times 10 = - 5 m s^{- 2}$
As $v^{2} = u^{2} + 2 a s,$
$∴ s = \frac{( v ^{2} - u ^{2} )}{2 a} = \frac{( 0 - ( 20 ) ^{2} )}{2 \times ( - 5 )} = 40 m$

Q. Consider a car moving along a straight horizontal road with a speed of 72kmh−1 . 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=10ms−2 )

30m

40m

72m

20m