A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s2 for 20 s. It then proceeds at constant speed for 1100 m before slowing down at 2.2 m/s2 until it stops at the station. A. What is the distance between the stations?
B. How much time does it take the train to go between the stations?

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330 wpu is be the right temperature

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.

Idon't  know if i got it wrong but i picked ( a )