SOLUTION: The current through a particular wire can be modeled by I(t) = 2t2 − 6t + 7 where I(t) is the current in amperes after t seconds. (a) Find the current in the wire afte

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The current through a particular wire can be modeled by I(t) = 2t2 − 6t + 7 where I(t) is the current in amperes after t seconds. (a) Find the current in the wire afte      Log On


   

Question 1025573: The current through a particular wire can be modeled by
I(t) = 2t2 − 6t + 7
where I(t) is the current in amperes after t seconds.
(a) Find the current in the wire after 0.5 seconds.
The current in the wire after 0.5 seconds would be about_____amperes.
(b) After how many seconds would the current reach 14 amperes? (Round your answer to two decimal places. Include units with your numerical answers.)
After about_____the current would reach 14 amperes.
(c) After how many seconds would the current reach 60 amperes? (Round your answer to two decimal places. Include units with your numerical answers.)
After about_____the current would reach 60 amperes.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The current through a particular wire can be modeled by
I%28t%29+=+2t%5E2-6t%2B7
where I%28t%29 is the current in amperes after t seconds.

This is another one of those "plug-numbers-into-the-equation" problems.

a) Substituting 0.5 for t , we get
I%280.5%29+=+2%2A0.5%5E2-6%2A0.5%2B7=2%2A0.25-3%2B7=0.5-3%2B7=4.5
The current in the wire after 0.5 seconds would be about highlight%284.5%29 amperes.

(b) I%28t%29=14 means 2t%5E2-6t%2B7=14
2t%5E2-6t%2B7=14-->2t%5E2-6t%2B7-14=0-->2t%5E2-6t-7=0 .
That quadratic equation can be solved by using the quadratic formula or by completing the square.
The quadratic formula says that the solutions to ax%5E2%2Bbx%2Bc=0 (if any)
are given by x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ .
In our case, system%28x=t%2Ca=2%2Cb=6%2Cc=-7%29 , so the solution comes from
=system%28t=about-0.9%2C%22or%22%2Ct=about3.8979%29.
The negative value for t makes no sense, because the problem does not tell us what happens for t%3C0; it is not part of the domain of the function; probably because t=0 was probably the moment the circuit was closed and current started flowing through the wire.
Rounding the positive answer to two decimal places, we get t=3.90 .
In part (a), we easily included units with your numerical answers, because the units (amperes) were already written for us after the blank space.
Here it is not so, but we know that t is measured in seconds and I%28t%29 is measured in amperes.
After about highlight%28%223.90+seconds%22%29 the current would reach 14 amperes.

(c) Similarly to what was done for (b), I%28t%29=60 means 2t%5E2-6t%2B7=60
2t%5E2-6t%2B7=60-->2t%5E2-6t%2B7-60=0-->2t%5E2-6t-53=0 .
Using the quadratic formula,
.
The positive solution, rounded to two decimal places is t=6.86 .
After about highlight%28%226.86+seconds%22%29 the current would reach 60 amperes.