The maximum velocity (in $m s^{- 1}$ ) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

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Solution

The correct option is B
30

$v = \sqrt{μ r g} = \sqrt{0.6 \times 150 \times 10} = 30 m / s$

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The maximum velocity (in $m s^{- 1}$ ) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

Q. The minimum velocity (in m/s) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is:

Q. The maximum velocity with which a driver must drive his car on a flat curved road of radius of curvature

150 m

and coefficient of friction

0.6

, to avoid the skidding off his car is (take

g = 10 m / s^{2}

)

The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m without skidding is $(a s s u m e g = 10 m s^{- 2})$

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The maximum velocity (in ms–1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

The correct option is B 30 v=√μrg=√0.6×150×10=30m/s

The maximum velocity (in $m s^{- 1}$ ) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

The correct option is B
30

$v = \sqrt{μ r g} = \sqrt{0.6 \times 150 \times 10} = 30 m / s$