What is the maximum acceleration of the passengers during this motion? What fraction is this of g?

Some passengers on an ocean cruise may suffer from motion sickness as the ship rocks back and forth on the waves. At one position on the ship, passengers experience a vertical motion of amplitude 1.4 m with a period of 15s .What is the maximum acceleration of the passengers during this motion? What fraction is this of g?Here, try this: divide your paper into three columns. In the first column put the information you are given in the problem. The first column looks like this: d = 1.4m; t = 15s; a = ?m/s^2.



In the middle column put the equation you will use to solve this problem. You know what equation to use by carefully studying the first column. Do you have an equation that has these variables in it: distance, time and acceleration? You should have found this equation: d = 陆 at^2. Put this equation in the middle column.



Now, this equation is the one we will use but first we need to use our algebra skills to solve this equation for 鈥榓鈥? This will give us a = 2d / t^2.



In the third column do the math work with the equation from the middle column and the information from the first column. All you have to do is replace the numbers with the symbols.



This will give you a = 2(1.4m) / (15s)^2. Do the math and see if you don鈥檛 get 0.012m/s^2.



To find what fraction of 鈥榞鈥?this is simply divide 鈥榞鈥?into this number.



Try solving other problems using this three-column method and see it this helps make the problems a lot easier to solve.What is the maximum acceleration of the passengers during this motion? What fraction is this of g?The maximum displacement is twice the amplitude or 2.8m and takes 7.5s. The average velocity is 2.8/7.5 m/s ( 0.373 m/s). The maximum velocity occurs at 0 displacement and assuming a constant acceleration would result in a maximum velocity of 0.75 m/s. Assuming a constant acceleration would result in an acceleration from 0 m/s to 0.75 m/s in one quarter period or 3.75 sec an acceleration of 0.2 m/s per s This approximately 2.0 % of g.What is the maximum acceleration of the passengers during this motion? What fraction is this of g?If we make some assumptions about the motion of the ship, this problem can be reduced to one of displacment under a constant acceleration. The displacement in this case will be 1.4 meters and it takes 7.5 seconds. The 15 seconds that is given would be for a complete wave; 7.5 seconds going up and then 7.5 seconds down.



1.4m = 1/2(a)(7.5sec)^2

1.4m(2)/(7.5sec)^2 = a = .0498m/sec2



(0.0498m/sec^2)/(9.81m/sec^2) = .005g



That shouldn't make anyone sick. Obviously, I've missed something.