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Katherine is speeding in her car, and sees a parked police car on the side of the road right next to her at t=0 seconds. She immediately decelerates, but the police car accelerates to catch up with her (assume the two cars are going in the same direction in parallel paths). The distance that Katherine has traveled in feet after t seconds can be modeled by the equation d(t) = 150 + 75t - 1.2t^2. The distance that the police car travels after t seconds can be modeled by the equation d(t) = 4t^2.a) How long will it take the police car to catch up to Katherine?b) How many feet has Katherine traveled from the time she saw the police car until the police car catches up to her?

Sagot :

When the police car catches up Katherine their distances from the point at t=0 are the same, then we can find the time that it takes them to advance that distance by making the expression that models the distance of Katherine and the equation for the distance of the police equal, like this:

[tex]\begin{gathered} 150+75t-1.2t^2=4t^2 \\ 150+75t-5.2t^2=0 \end{gathered}[/tex]

We know that for an equation of the form at^2+bt+c=0 the value of t can be calculated with the quadratic formula, for this case a= -5.2, b=75 and c=150, then applying the quadratic formula we get:

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