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Solve the Problem ft\mathrm { ft } ) of the Rocket Above the Ground (The Position Function)

Question 150

Multiple Choice

Solve the problem.
-A rocket is launched straight up from level ground. The distance (in ft\mathrm { ft } ) of the rocket above the ground (the position function) is f(t) =224t16t2\mathrm { f } ( \mathrm { t } ) = 224 \mathrm { t } - 16 \mathrm { t } ^ { 2 } at any time t\mathrm { t } (in sec) . Find f(3) \mathrm { f } ^ { \prime } ( 3 ) and the initial velocity of the rocket.


A) f(3) =224\mathrm { f } ^ { \prime } ( 3 ) = 224 ; initial velocity =128ft/s= 128 \mathrm { ft } / \mathrm { s }
B) f(3) =128\mathrm { f } ^ { \prime } ( 3 ) = 128 ; initial velocity =224ft/s= 224 \mathrm { ft } / \mathrm { s }
C) f(3) =528\mathrm { f } ^ { \prime } ( 3 ) = 528 ; initial velocity =224ft/s= 224 \mathrm { ft } / \mathrm { s }
D) f(3) =48f ^ { \prime } ( 3 ) = - 48 ; initial velocity =224ft/s= 224 \mathrm { ft } / \mathrm { s }

Correct Answer:

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