The formula for braking distance of a car:
d=0.018268 * v^2/0
where v0 is the initial velocity of the car (feet per second) and d is the braking distance (ft). In this model, the roadway grade is kept constant at 5% and the coefficient of friction at 0.8. In a school zone, you want the maximum breaking distance to be 10 feet since this seems like a reasonable distance to see a child who might be in the way of a driver.

A)What should you set as the speed limit (i. e. initial velocity of the car) so that the breaking distance is 10 feet or less? Round the the nearest foot per second.
feet per second.

B). The equation x^2=9 has two solutions: 3 and -3. With that in mind, what is the other solution to the equation 10=0.018268⋅v^20, besides what you answered in (a)? Why is this other solution not the correct answer to the question in (a)?
The second solution is feet per second.

show us the graphs dude

step-by-step explanation:

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation:

answer: i think you should do the math and pay attention in class : )

step-by-step explanation:

that is truly what   think but if i get the answer i will