At a certain point in the reentry of the space shuttle into the earth's atmosphere, the total acceleration of the shuttle may be represented by two components. One component is the gravitational acceleration g = 9.63 m/s at this altitude. The second component equals 10.78 m/s due to atmospheric resistance and is directed opposite to the velocity. The shuttle is at an altitude of 45.1 km and has reduced its orbital velocity of 28300 km/h to 16190 km/h in the direction = 1.76. For this instant, calculate the radius of curvature of the path and the rate i at which the speed is changing. Answers: p= i = i 330.170 i 18.753 km m/s