An object is traveling at a constant velocity of 15m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11's. what will its velocity after that acceleration?

That all depends on the direction of the 15 m/s and the direction of the acceleration. You've only given us a speed and a magnitude of acceleration, with no directions. So we have no initial velocity, and no way to calculate a final one. / / / IF everything is in the same direction, then the final speed in that same direction is (15 m/s) + (3.5 m/s^2 x 11 sec). That's (15 m/s) + (38.5 m/s) = 53.5 m/s.

Ineed the problem dude

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.