Respuesta :
Assume uniformly accelerated motion
Vf = Vo - at => a= [Vo - Vf] / t
Vf = 0, t = 20 s => a = Vo / 20
x = Vot - at^2 / 2
x = Vot - (Vo / 20) * t^2 /2
x = 1775 feet * [1 mile / 5280 feet] = 0.336 mile
t = 20 s
=> 0.336 = 20Vo - Vo(20)^2 /(40) = 20Vo - 10 Vo = 10 Vo
=> Vo = 0.336 /10 = 0.0336 mile /s
Vo = 0.0336 mile/s * 3600 s/h = 120.96 mile / h
Therefore, the car was way over the speed limit.
Vf = Vo - at => a= [Vo - Vf] / t
Vf = 0, t = 20 s => a = Vo / 20
x = Vot - at^2 / 2
x = Vot - (Vo / 20) * t^2 /2
x = 1775 feet * [1 mile / 5280 feet] = 0.336 mile
t = 20 s
=> 0.336 = 20Vo - Vo(20)^2 /(40) = 20Vo - 10 Vo = 10 Vo
=> Vo = 0.336 /10 = 0.0336 mile /s
Vo = 0.0336 mile/s * 3600 s/h = 120.96 mile / h
Therefore, the car was way over the speed limit.
When the driver first pressed on the brake to stop the car, the car was traveling at speed of 59.17 feet/second at that point
Let us assume
Initial velocity =u
Final velocity = v
What is acceleration?
It is the rate of change of velocity in a given time period.
So acceleration [tex]a = \frac{v-u}{t}[/tex]
Where v=0 because after pressing the brake, the car comes to the rest.
t = 20 seconds
[tex]a = \frac{-u}{20}[/tex]
As we know that
distance [tex]s= ut+\frac{1}{2} *a*t^2[/tex]
s =1775 feet(given)
put the value [tex]a = \frac{-u}{20}[/tex] in the above equation
[tex]1775= 20u+\frac{1}{2} * \frac{-u}{20} *20^2[/tex]
[tex]u = 59.17 feet/second[/tex]
Hence, When the driver first pressed on the brake to stop the car, the car was traveling at speed of 59.17 feet/second at that point.
To get more about acceleration visit to:
https://brainly.com/question/950219