Car a is traveling at 22.0 m/s and car b at 29.0 m/s. car a is behind car b when the driver of car a accelerates his car with an acceleration of 2.40 m/s2 . it takes car a 19.0 s to catch up to car b. how far ahead of car a was car b, when car a began to accelerate?

Answerapply an acl that has all deny ace statement;

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.

D) magnetic attraction

Answer to the left (eastward); to the right (westward)