A rolling cart carrying books has a mass of 30 kg and is shoved so that it is given an initial velocity of 3 m/s. The cart's bearings are such that the coefficient of friction can be treated as 0.2 and friction acts to slow the cart to rest. a) What is the magnitude of the cart's acceleration as friction acts upon it?
b) How much time does it take for the cart to come to rest?
c) How far does the cart roll before it comes to rest?

1. may 2. september 3. february, and december4. 70°f

Special care would be an answer

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.