17–22. position from velocity consider an object moving along a line with the given velocity v and initial position. a. determine the position function, for t ú 0, using the antiderivative method b. determine the position function, for t ú 0, using the fundamental theorem of calculus (theorem 6.1). check for agreement with the answer to part (a). 17. v1t2 = sin t on 30, 2p4; s102 = 1

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Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.