A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as

Question

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 19.2t and y = 3.60t ? 5.29t2, where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball's position as a function of time, using the unit vectors and . (Give the answer in terms of t.) vector "r" = By taking derivatives, do the following. (Give the answers in terms of t.) (b) obtain the expression for the velocity vector as a function of time vector "v" = (c) obtain the expression for the acceleration vector as a function of time vector "a"= (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.06 s. vector "r" = m vector "v"= m/s vector "a = m/s2

Explanation / Answer

a. r = (19.2t)mi + (3.60t -5.29t^2)mj b. v = (19.2)m/si + (3.60-10.58t)m/sj c. a = (0)m/s^2i + (-10.58)m/s^2j d. r = 58.752mi - 38.517444 mj v = 19.2m/si -28.7748m/sj c = 0m/s^2 i - 10.58m/s^2j The d values were gotten from just plugging in t= 3.06. Sorry I was not able to answer fast enough

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