Question

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°?

Answer 1

83.1946504051 m

Step-by-step explanation:

u = Initial velocity =
`60\ mph=(60* 1609.34)/(3600)=26.82233\ m/s`

s = Displacement =
`123\ ft=(123)/(3.281)=37.4885705578\ m`

`\theta` = Angle =
`26^(\circ)`

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(0^2-26.82233^2)/(2* 37.4885705578)\\\Rightarrow a=-9.5954230306\ m/s^2`

Coefficient of friction

`\mu=-(a)/(g)\\\Rightarrow \mu=(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=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-26.82233^2)/(2* -4.32382)\\\Rightarrow s=83.1946504051\ m`

The stopping distance is 83.1946504051 m