. A car is traveling at 30 m/s and about to impact on a wall. What change would result in the safest outcome for the passengers in the car?

A large time of impact with the wall

A small time of impact with the wall

A large stopping force with the wall

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18

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.