What will be the maximum speed of a car on a road-turn of radius 30 m if the coefficient of friction between the tyres and the road is 0.4? (Take g=9.8 m s- 2 )

Question

What will be the maximum speed of a car on a road-turn of radius 30 m if the coefficient of friction between the tyres and the road is 0.4? (Take g=9.8 m s- 2 ) (A) 10.84ms- 1 (B) 9.84ms- 1 (C) 8.84ms- 1 (D) 6.84ms- 1

Tardigrade Admin · Accepted Answer

Given r = 30 m and μ = 0.4. vmax = √μ r g = √0.4 × 30 × 9.8 = 10.84 m s- 1 For a safe turn (M v2/r)≤ μ s Mg or v2≤ μ s rg where μ s arrow coefficient of static friction

Q. What will be the maximum speed of a car on a road-turn of radius $30 m$ if the coefficient of friction between the tyres and the road is $0.4 ?$ (Take $g = 9.8 m s^{- 2}$ )

$10.84 m s^{- 1}$

$9.84 m s^{- 1}$

$8.84 m s^{- 1}$

$6.84 m s^{- 1}$

Given
$r = 30 m an d μ = 0.4.$
$v_{ma x} = μ r g = 0.4 \times 30 \times 9.8 = 10.84 m s^{- 1}$
For a safe turn
$\frac{M v ^{2}}{r} \leq μ_{s} M g or v^{2} \leq μ_{s} r g$
where $μ_{s} \to$ coefficient of static friction

Q. What will be the maximum speed of a car on a road-turn of radius 30m if the coefficient of friction between the tyres and the road is 0.4? (Take g=9.8ms−2 )

10.84ms−1

9.84ms−1

8.84ms−1

6.84ms−1