i have all of the following formulas: vf = vi+at d = (vf+vi)/2 x t d = (vi)(t) + 1/2(a)(t)^2 vf^2 = vi^2 + 2(a)(d) 1) Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a safety bel can withstand an acceleration of -300 m/s. how thick should barriers be to safety stop a car that hits a barrier at 110 km/h? First, I converted 110 km/h to 30.6 m/s. Given: A = -300 m/s Vi = 30.6 m/s Vf = 0 m/s Find: d Solution: i used vf^2 = vi^2 + 2(a)(d) and plugged in all of the numbers given and solved for d. vf^2 = vi^2 + 2(a)(d) (0 m/s^2) = (30.6 m/s)^2 + 2(-300 m/s-squared)(d) -936.36 m/s = 2(-300 m/s-squared)(d) -936.36 m/s = (-600 m/s)(d) d = 1.56m 2) A baseball pitcher throws a fastball at a speed of 44 m/s. The acceleration occurs as the pitcher holds the ball in his hand and moves it through an almost straight line distance of 3.5 m. Calculate the acceleration, assuming it is uniform. Compare this acceleration to the acceleration due to gravity, 9.80 m/s-squared. Given: Vi = 0 m/s Vf = 44 m/s d = 3.5 m Find: a Solution: vf^2 = vi^2 + 2(a)(d) 44 m/s^2 = 0 m/s^2 + 2(a)(3.5m) 1936 m/s = (7m)(a) a = 276.57 m/s-squared compared to gravity: 276.57 m/s-squared / 9.80 m/s-squared = 28.22 3) If a bullet leaves the muzzle of a rifle with a speed of 600 m/s, and the barrel of the rifle is 0.9m long, what is the acceleration of the bullet while in the barrel? Given: Vi = 0 m/s Vf = 600 m/s d = 0.9 m Find: a Solution: vf^2 = vi^2 + 2(a)(d) 600 m/s^2 = 0 m/s^2 + 2(a)(0.9m) 360000 m/s = (1.8m)(a) a = 200000 m/s-squared thanks!