Explanation Given information: The provided expression: s ( t ) = − 16 t 2 + 72 t . And, the time is 3 seconds, Formula Used: If a function expresses an object’s position in terms of time, then the instantaneous velocity at time t = a is given by, s ' ( x ) = lim h → 0 s ( a + h ) − s ( a ) h Calculation: Consider the expression, s ( t ) = − 16 t 2 + 72 t . To calculate the derivative substitute the values in the above formula as shown below: s ' ( t ) = lim h → 0 − 16 ( t + h ) 2 + 72 ( t + h ) − ( − 16 t 2 + 72 t ) h = lim h → 0 − 16 ( t 2 + h 2 + 2 t h ) + 72 t + 72 h + 16 t 2 <

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Chapter 11, Problem 16T

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To Calculate: The instantaneous velocity after 3 seconds of a ball is thrown where height above the ground is given by s(t)=−16t2+72t.

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Chapter 11, Problem 15T

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Chapter 11 Solutions

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Ch. 11.1 - Check Point 1 Find: . Ch. 11.1 - Prob. 2CP Ch. 11.1 - Prob. 3CP Ch. 11.1 - Prob. 4CP Ch. 11.1 - Prob. 5CP Ch. 11.1 - Prob. 1CVC Ch. 11.1 - Prob. 2CVC Ch. 11.1 - Prob. 3CVC Ch. 11.1 - Fill in each blank so that the resulting statement...Ch. 11.1 - Fill in each blank so that the resulting statement...

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The instantaneous velocity after 3 seconds of a ball is thrown where height above the ground is given by s ( t ) = − 16 t 2 + 72 t .

Solution Summary: The author calculates the instantaneous velocity after 3 seconds of a ball is thrown where height above the ground is given by underset_-24ft/sec

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To Calculate: The instantaneous velocity after 3 seconds of a ball is thrown where height above the ground is given by s(t)=−16t2+72t.

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Chapter 11 Solutions