SOLUTION: A bug is placed on the road and its distance t seconds after it is put down is p(t)= -16t^2 + 64t. At what moment will the bug be 14 feet from where it began? At what time will th

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Question 604361: A bug is placed on the road and its distance t seconds after it is put down is p(t)= -16t^2 + 64t. At what moment will the bug be 14 feet from where it began?
At what time will the bug turn around and head back to its starting place?
How far from where it began will the bug be when it turns around?
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A bug is placed on the road and its distance t seconds after it is put down is p(t)= -16t^2 + 64t.
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At what moment will the bug be 14 feet from where it began?
p(t)= -16t^2 + 64t = 14
-16t^2 + 64t - 14 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=3200 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 0.232233047033631, 3.76776695296637. Here's your graph:

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At these 2 times, once going, once returning.
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At what time will the bug turn around and head back to its starting place?
p(t)= -16t^2 + 64t
p'(t) = -32t + 64 = 0
t = 2 seconds
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How far from where it began will the bug be when it turns around?
p(2) = -16*4 + 64*2
= 64 feet
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