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4. The position of a point on a line is given by the equation s (t) = t3 - 6t + 9t - 4, where

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4. The position of a point on a line is given by the equation s (t) = t3 - 6t + 9t - 4, where s is measured in metres and t in seconds. (2 marks each) a. What is the velocity of the point after 2 seconds? b. What is its acceleration after 4 seconds? c. Where is it when it first stops moving? d. How far has it travelled when its acceleration is 0? e. After 2 second, is it moving toward or away from the origin?1. Find the derivatives of each of these functions. (3 marks each) a. y = (2 + 7)3 (2 - 9)4 b . y = 202 -5 x - 5 c. y = (2 + 4 4x2 + 2x - 5 d. y = V/2x2+5 ( 20 + 3 ) 4 2. Use the process of implicit differentiation to find ~ given that: 5x y+ + % = 5xy. Apply implicit differentiation directly without first simplifying the equation. (10 marks) 3. Give an example of a situation in which composite differentiation might be used. Give examples of functions that might be applicable in your situation, and show how the relevant rates of change might be calculated. (8 marks)

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