Asked by Yoona
The three pigls are building their houses while the brik=ck house building pig and the stick house building pig are working at a steady rate, the straw house building pig(knowing the wolf will arrive first at his house) is frantically increasing his rate of building by 2feet/minute. the straw house building pig has to make his house 16 ft tall(right now, it's 0 feet tall) and the wolf is scheduled to arrive in 5 minutes. Will the pig have his house built for the wolf to blow down by the time he arrives?
Answers
Answered by
naseba
are you sure this is a question?
Answered by
Steve
this is just a standard acceleration problem.
s = 1/2 at^2
16 = 1/2 * 2 * t^2
16=t^2
t=4
so, in 4 min the house will be 16' tall, and he will be (briefly) safe.
s = 1/2 at^2
16 = 1/2 * 2 * t^2
16=t^2
t=4
so, in 4 min the house will be 16' tall, and he will be (briefly) safe.
Answered by
Steve
now, if you want to use your calculus, let f(t) be the height of the house after t minutes.
f ''(t) = 2
f '(t) = 2t + c
f '(0) = 0 means c=0 and
f '(t) = 2t
f(t) = t^2+c
f(0) = 0 means c=0, so
f(t) = t^2
16 = t^2
t=4
f ''(t) = 2
f '(t) = 2t + c
f '(0) = 0 means c=0 and
f '(t) = 2t
f(t) = t^2+c
f(0) = 0 means c=0, so
f(t) = t^2
16 = t^2
t=4
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