The Nuclear Missile Crisis

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  1. The Nuclear Missile Crisis Information brought to you by CIA mathematicians; Taylor Flinchbaugh, Jenna Hammers, Christina Hayes, and Melanie Otte 2.…
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  • 1. The Nuclear Missile Crisis Information brought to you by CIA mathematicians; Taylor Flinchbaugh, Jenna Hammers, Christina Hayes, and Melanie Otte
  • 2. <ul><li>We have launched a defensive missile to block the attack. </li></ul><ul><li>Will our missile be able to block the nuclear warhead in time to save the moon? </li></ul>
  • 3. <ul><li>First, we used the total distance traveled by each missile each day beginning with the North Korean Missile and put the data into a stat plot. The day being the x value in the first column and the y value the distance traveled from Earth in miles. </li></ul>
  • 4. <ul><li>Then we graphed the points by pressing [zoom] [9] to fit the data for the North Korean Missile. </li></ul><ul><li>We found the line of best fit by pressing [STAT] -> [CALC] [4] [2 nd ] [1] [,] [2 nd ] [2] [ENTER] </li></ul><ul><li>We used the a and b values to form the equation for the line of best fit for the path of the missile. </li></ul><ul><li>y= 24388.05952x – 21474.39286 </li></ul>
  • 5. <ul><li>We found the distance travelled each day by subtracting the distance from the previous day from the current day for each day. We made sure to include the first day’s distance since we didn’t have to subtract that from anything. </li></ul><ul><li>We added all distances per day together and divided by 8 to get the average distance travelled per day. </li></ul>
  • 6. <ul><li>The average miles per day = 22,081.88 </li></ul><ul><li>To get the km/ day we multiplied the average miles/ day by 1.609 (conversion factor, 1 mile= 1.609 km) </li></ul><ul><li>km/day = 35,529.74 </li></ul><ul><li>To get the average miles and km per hour we divided the average miles and km per day by 24 (conversion factor, 1 day= 24 hrs) </li></ul><ul><li>mph = 22,081.88/24 </li></ul><ul><li>mph = 920.08 </li></ul><ul><li>km/hr = 35,529.74/24 </li></ul><ul><li>km/hr = 1,480.41 </li></ul>
  • 7. United States’ Counter Missile <ul><li>We repeated the same steps for the American Missile as we did for the North Korean Missile. </li></ul><ul><li>We plotted the points in a stat plot, graphed and found the line of best fit. </li></ul><ul><li>The equation for the line of best of the path of the American missile is </li></ul><ul><li>Y=26176.40476x - 40085.07143 </li></ul>
  • 8. <ul><li>We found the distance travelled each day by subtracting the distance from the previous day from the current day for each day. We made sure to include the first day’s distance since we didn’t have to subtract that from anything. </li></ul><ul><li>We added all distances per day together and divided by 8 to get the average distance travelled per day. </li></ul>
  • 9. <ul><li>The average miles per day = 21,632.13 </li></ul><ul><li>To get the km/ day we multiplied the average miles/ day by 1.609 (conversion factor, 1 mile= 1.609 km) </li></ul><ul><li>km/day = 34,806.10 </li></ul><ul><li>To get the average miles and km per hour we divided the average miles and km per day by 24 (conversion factor, 1 day= 24 hrs) </li></ul><ul><li>mph = 21,632.13 /24 </li></ul><ul><li>mph = 901.34 </li></ul><ul><li>km/hr = 34,806.10/24 </li></ul><ul><li>km/hr = 1,450.25 </li></ul>
  • 10. <ul><li>Then by going to the [y=] screen we typed in both equations of best fit lines. The North Korean missile’s equation in Y1 and the American missile’s equation in Y2. </li></ul><ul><li>Press [GRAPH] </li></ul><ul><li>Press [ ZOOM] [3] to zoom out. </li></ul>
  • 11. <ul><li>While on the graph screen we pressed [2 nd ] [TRACE] [5] </li></ul><ul><li>Then we pressed [Enter] three times, then the intersection points appeared on the bottom of the screen. </li></ul><ul><li>X=10.406648 </li></ul><ul><li>Y= 232323.55 </li></ul><ul><li>This means that on day 10.41 (10 days and about 10 hours) at a distance of 232,323.55 miles from Earth the missiles will intersect. </li></ul>
  • 12. <ul><li>y = 24388.05925x - 21474.39286 </li></ul><ul><li>y = 26176.40476 x - 40085.07143 </li></ul><ul><li>24388.05925x - 21474.39286 = 26176.40476x - 40085.07143 </li></ul><ul><li> + 40085.07143 + 40085.07143 </li></ul><ul><li>24388.05925x + 18610.67857 = 26176.40476x </li></ul><ul><li>-24388.05925x -24388.05925x </li></ul><ul><li>18610.67857 = 1788.34551x </li></ul><ul><li>/1788.34551 /1788.34551 </li></ul><ul><li>x = 10.40664596 </li></ul><ul><li>y = 24388.05925(10.40664596)-21474.39286 </li></ul><ul><li>y = 232323.5054 </li></ul><ul><li>Using substitution we found that the missiles will intersect at 10.41 (10 and about 10 hours) days at 232,323.51 miles away from Earth. </li></ul>
  • 13. Will the Moon be spared? <ul><li>The moon is approximately 238,897 miles away (average distance from center of Earth to center of the moon). </li></ul><ul><li>According to calculations the missile intersection will be at 232,323.55 miles from the Earth, therefore the Moon will be spared. </li></ul>
  • 14. Possible Errors <ul><li>The lines of best fit aren’t the exact paths of the missiles, they are estimations, therefore the intersection point is most likely slightly off. </li></ul><ul><li>The missiles didn’t travel at a constant rate each day therefore their path before their intersection is unknown, it can only be estimated. </li></ul><ul><li>The distance from the Earth’s surface to the moon is slightly different for each missile since its take-off point was different. Also, the measurement used was from the center of the Earth to the center of the Moon, not from surface to surface so the distance from Earth will be slightly different. </li></ul>
  • 15. Roles <ul><li>This project was a group effort, we worked on everything together. </li></ul>
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