18) First calculate the retreat rate (i.e., velocity) in feet per year at both p

Question

18) First calculate the retreat rate (i.e., velocity) in feet per year at both points X and Y, and then average your results. Remember that velocity equals distance divided by time. retreat rate at X: retreat rate at Y: average retreat rate: 19) Using the average rate of retreat, predict the number of years it will take for the landward side of Assateague Island at point Z to reach the shore at point Z'. Again, use the formula of velocity equals distance over time. distance between Z and Z' travel time at Z: 20) Based on the number of years you calculated above, determine what year (e.g., 1999) Assateague Island should have reached the mainland, sealing off Sinepuxent Bay.
Figure 7.5- (USGS Ocean City, MD Ocean City WETT 2 1933 of CEAN O Ex 7-Coastal Hazards

Explanation / Answer

18. Number of years = 1972 - 1933 = 39

Retreat rate at X = ( Distance between X and X' ) / (Number of years) = 1788 / 39 = 45.84 feet / year, as the distance on the map is 1.2 cm equivalent to 545 m, i.e. 1788 feet

Similarly, Retreat rate at Y =  2385 / 39 = 61.15 feet/year since according to scale,1.6 cm on the map is equal to 727 m, i.e. 2385 feet

Average retreat rate = (61.15 + 45.84) / 2 = 53.49 feet / year

18. The distance between Z and Z' = 1191 feet, as on the map the distance is 0.8 cm, i.e. 363 m on the earth's surface.

Using the average retreat rate of 53.49 feet/year, we find that point Z will reach point Z' in

1191 / 53.49 = 22 years (approx)

20. Adding 22 years since 1972, we find that the Assateague Island should have reached the mainland by the year 1994.

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