Question

an automobile is travelling at a speed of 54km/hr . if the coefficient of static friction between tyres and the road is 0.60 . calculate the minimum distance in which the vehicle would be stopped​

Answer 1

Answer:

an automobile is traveling at a speed of 54 km/h if the coefficient of static friction between the tyres and the road is 0.60 , calculate the minimum distance in which the vehicle would be stopped

Explanation:

Answer 2

The minimum distance in which the vehicle would be stopped is 18.75 m.

Given:

Initial speed of automobile (u) = 54 km/hr

Coefficient of static friction = 0.60

To Find:

The minimum distance in which the vehicle stops

Solution:

Let the minimum distance in which the vehicle stops be 'S'.

The initial speed of the vehicle (u in km/hr) = 54 km/hr

                             $u (m/s)= (54)(\frac{5}{18})=15(m/s)$

The initial speed of the vehicle (u in ms⁻¹) = 15 ms⁻¹

Coefficient of static friction (μ) = 0.60

Frictional force on vehicle (F) = μmg

                                   $Retardation = \frac{Friction}{Mass}=\frac{F}{m}$

The retardation (a) suffered by the vehicle due to friction = μg

Assumption: g = 10 ms⁻²

∴ a = μg = (0.60) × (10) = 6 ms⁻²

→ a = 6 ms⁻²

We can now use the 3rd equation of motion: v² = u² + 2aS where the final speed (v) would be zero as the vehicle finally stops.

→ v² = u² + 2aS

→ 0² = (15)² + 2(-6)(S)

→0 = 225 - 12S

 $S=\frac{225}{12} =18.75m$

∴ S = 18.75 m

Hence, the minimum distance in which the vehicle would be stopped is 18.75 m.

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