A woman attached to a bungee cord jumps from a bridge that is 2626 m above a river. Her height in meters above the river t seconds after the jump is ​y(t)equals=13 left parenthesis 1 plus e Superscript negative t Baseline cosine t right parenthesis131+e−tcost​, for tgreater than or equals≥0. a. Determine her velocity at tequals=1 and tequals=33. b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 s. c. Use a graphing utility to estimate the maximum upward velocity. Question content area bottom Part 1 a. Her velocity at time t is given by the function ​v(t)equals=negative 13 e Superscript negative t Baseline left parenthesis sine t plus cosine t right parenthesis−13e−t(sint+cost). Part 2 Her velocity at tequals=1 is negative 6.61−6.61. ​(Round to two decimal places as​ needed.) Part 3 Her velocity at tequals=33 is 0.550.55. ​(Round to two decimal places as​ needed.) Part 4 b. Choose the correct answer below. A. She is moving downward for 1.571.57less than<tless than<4.714.71 and 7.857.85less than<tless than or equals≤10. She is moving upward for 0less than or equals≤tless than<1.571.57 and 4.714.71less than<tless than<7.857.85. B. She is moving downward for 0less than or equals≤tless than<2.362.36 and 5.505.50less than<tless than<8.648.64. She is moving upward for 2.362.36less than<tless than<5.505.50 and 8.648.64less than<tless than or equals≤10. C. She is moving downward for 0.790.79less than<tless than<3.933.93 and 7.077.07less than<tless than or equals≤10. She is moving upward for 0less than or equals≤tless than<0.790.79 and 3.933.93less than<tless than<7.077.07. D. She is moving downward for 3.143.14less than<tless than<6.286.28 and 9.429.42less than<tless than or equals≤10. She is moving upward for 0less than or equals≤tless than<3.143.14 and 9.429.42less than<tless than or equals≤10.