A spaceship ferrying workers to moon-base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Suppose the spaceship starts from rest and
accelerates at 20 m/s2
, for the first 15 min of the trip, and then travels at constant velocity
until the last 15 min when it slows down at a rate of 20 m/s2
, just coming to rest as it
reaches the moon.
a) What is the maximum velocity attained? (Ans: 1.8 x 104
m/s)
b) What fraction of the total distance is traveled at constant speed? ( Ans: 95.78 % )
c) What total time is required for the whole trip?

A) v₁ = a₁t ₁ = 19.9\*14.9\*60 = 1.78e4 m/s b) Distance traveled with acceleration = 2\* (v₁/2)\*t₁= v₁ t ₁ = 1.78e4 \*14.9\*60 = 1.59e7 m Distance traveled without acceleration 3.84e8 -1.59e7 = 3.68e8m The fraction is 3.68e+8m / 3.84e8 = 96% c) Time traveled without acceleration = 3.68e8/(60\*1.78e4) = 344.57 minutes Total time = 344.57 + 2\*14.9 =374.37 minutes"

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.