Asked by sara
                The equation d=0.12v^2 + 2.1v models the distance d in feet it takes a car traveling at a speed of v miles per hour to come to a complete stop. Graces car was able to stop after 350ft on a highway. If the speed limit in the area was 35 miles per hour, was she speeding? Justify your answer.
I did 350=0.12v^2 +2.1v
-350to both sides -350
0=0.12v^2 +2.1v-350
a=.12
b=2.1
c=-350
x=-b+- square root b^2-4ac/2a
x=-2.1+- square root 4.41-4(.12)(-350)/2(.12)
x=-2.1+- square root172.41/.24
square root of 172.41= 13.13, now divide that by .24=54.71
-2.1 +54.71 = 52.61
-2.1-54.71= -56.81 - we don't use this one
So if the speed limit was 35 mph , Grace was speeding, going approx. 52 mph
Can you please check my work? Is this correct?
Thank you for your help!!
            
            
        I did 350=0.12v^2 +2.1v
-350to both sides -350
0=0.12v^2 +2.1v-350
a=.12
b=2.1
c=-350
x=-b+- square root b^2-4ac/2a
x=-2.1+- square root 4.41-4(.12)(-350)/2(.12)
x=-2.1+- square root172.41/.24
square root of 172.41= 13.13, now divide that by .24=54.71
-2.1 +54.71 = 52.61
-2.1-54.71= -56.81 - we don't use this one
So if the speed limit was 35 mph , Grace was speeding, going approx. 52 mph
Can you please check my work? Is this correct?
Thank you for your help!!
Answers
                    Answered by
            Nonetheless
            
    It looks good most of the way; just a small error near the end:
First work out -2.1 + 13.13, and then divide the result by 0.24.
Same correction needed for your second solution.
    
First work out -2.1 + 13.13, and then divide the result by 0.24.
Same correction needed for your second solution.
                    Answered by
            sara
            
    thank you
so I got 45.95, so approx. 46mph, which is still over the speed limit.
Thank you for taking the time to check my long involved problem.
    
so I got 45.95, so approx. 46mph, which is still over the speed limit.
Thank you for taking the time to check my long involved problem.
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