In the aftermath of a car accident, it is concluded that one driver slowed to a halt in 20 seconds while skidding 1775 feet. If the speed limit was 60 miles per hour, can it be proved that the driver had been speeding? (Hint: 60 miles per hour is equal to 88 feet per second.) We can guarantee that at some time from when the driver first pressed on the brake to when the car came to a complete stop the car was traveling

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.



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.

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