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A race car driving under the caution flag at 80 feet per second begins to accelerate at a constant rate after the warning flag. The distance traveled since the warning flag in feet is characterized by 30t 2 + 80t, where t is the time in seconds after the car starts accelerating again.How long does it take the car to travel 30 feet after it begins accelerating?

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Answer: 1/3 sec or 0.333 sec


Step-by-step explanation:


d = 30*t^2 + 80*t = 30

putting into quadratic form, a*t^2 + b*t + c = 0

30*t^2 + 80*t - 30 = 0 where: a = 30 b = 80 c = -30

solve quadratic:

t = [-b + sqrt( b^2 - 4*a*c)]/2*a

substitute in values above

therefore, t = 1/3 sec or 0.333 sec

Cetacea

It takes [tex]\frac{1}{3}[/tex] seconds to travel 30 feet after it begins accelerating.

Given that: The distance traveled since the warning flag in feet is characterized by [tex]30t ^2 + 80t[/tex].

Given that: the car travelled 30 feet after it begins accelerating.

It means d = 30.

[tex]30t^2 + 80t=30\\30t^2+80t-30=0\\30t^2+90t-10t-30=0\\30t(t+3)-10(t+3)=0\\(t+3)(30t-10)=0\\t=-3,\frac{1}{3}[/tex]

Time(t) can not be negative. So, [tex]t=\frac{1}{3}[/tex].

Learn more: https://brainly.com/question/24396559

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