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damaskus
15 days ago
8

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

is its stopping distance on a roadway sloping downward at an angle of 26.0°?
Physics
1 answer:
kicyunya [1K]15 days ago
6 0

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

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