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The coefficient of friction between a ca...

The coefficient of friction between a car wheels and a roadway is 0.5 The least distance in which the car can accelerate from rest to a speed of 72 kmph is `(g=10 ms^(-2))`

A

10m

B

20m

C

30m

D

40m

Text Solution

Verified by Experts

The correct Answer is:
D
`v^(2)-u^(2)=2as, a=mu_(k)g`
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Knowledge Check

  • Consider a car moving along a straight horizontal road with a speed of 72 km / h . If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is [g=10ms^(-2)]

    A
    30 m
    B
    40 m
    C
    72 m
    D
    20 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 [g=10 ms^(-1)]

    A
    20m
    B
    40m
    C
    30m
    D
    72m

  • Consider a car moving along a straight horizontal road with a speed of 36 km/h. If the coefficient of static friction between road and tyers is 0.4, the shortest distance in which the car can be stopped is (Take g=10 m//s2)

    A
    `33.8 m`
    B
    `12.5 m`
    C
    `58.6 m`
    D
    `20m`

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