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A rock thrown vertically upward from the surface of the moon at a velocity of 40 mlsec reaches a height of s = 40t 0.8t2

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A rock thrown vertically upward from the surface of the moon at a velocity of 40 mlsec reaches a height of s = 40t 0.8t2 meters in t sec. a. Find the rock's velocity and acceleration at time t. b. How long does it take the rock to reach its highest point? c. How high does the rock go? d. How long does it take the rock to reach half its maximum height? 9. How long is the rock aloft? a. Find the rock's velocity at time t. v=|: mils Find the rock's acceleration at time t. a= :l mis2 (Simplify your answer. Use integers or decimals for any numbers in the expression.) b. How long does it take the rock to reach its highest point? sec (Simplify your answer.) c. How high does the rock go? m (Simplify your answer.) d. How long does it take the rock to reach half its maximum height? sec (Simplify your answer. Round to two decimal places as needed. Use a comma to separate answers as needed.) 9. How long is the rock aloft? sec (Simplify your answer.) When a circular plate of metal is heated in an oven, its radius increases at a rate of 0.05 cm/min. At what rate is the plate's area increasing when the radius is 51 cm? . . The rate of change of the area is cm2/min. (Type an exact answer in terms of It.)Given .. radius increases at a rate of 0.05 cm/ min using the formula for the area of circle . A = TIX 2 where of is the area of the circle and I t's radius. To find the rate at which the area is incusing , we can differentiate Both sides of the equation with respect to time t: LA de where dA Is the rate at which the area is increasing and lolcat is the rate at which radius is incusing . We are given that the ralives is increasing at a rate of 0.05 cm/ min , Do we can substitute this into the equation: da = 271 (51 ) (0.05) simplifying this , we get : dA 16. 02 Em? / min The rate of change of the area is 16 .02 1 cm 2 / min

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