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purplemexican

  • one year ago

On the Moon, a falling object falls just 2.65 feet in the first second after being dropped. Each second it falls 5.3 feet farther than in the previous second. How far would an object fall in the first ten seconds after being dropped?

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  1. welshfella
    • one year ago
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    this is a an arithmetic series where the first term is 2.65 and each subsequent term is obtained by adding the common difference 35.

  2. Purplemexican
    • one year ago
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    a. 270 ft c. 313 ft b. 692 ft d. 265 ft

  3. welshfella
    • one year ago
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    nth term = a1 + ( n- 1)d where a1 = first term , n = number of terms and d = common difference so tenth term = 2.65 + 9*5.3 (common difference = 5.3 NOT 35 as i said in my first post)

  4. welshfella
    • one year ago
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    so in the second second it traveled 2.65 + 5.3 = 7.95 feet the total distance = sum of n terms = (n/2)[ a1 + L) where L = 10th term so plug in the values of a1 , n (=10) , and L and work it out

  5. Purplemexican
    • one year ago
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    what the formula

  6. welshfella
    • one year ago
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    i've just given it

  7. Purplemexican
    • one year ago
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    im confused please state it again if you don't mind

  8. welshfella
    • one year ago
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    plug in a1 = 2.65 , n = 10 and L = 10th term = 2.65 + 9*5.3 = 50.35

  9. Purplemexican
    • one year ago
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    so L=50.35 correct?

  10. welshfella
    • one year ago
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    thats the distance travelled in the 10 th second . You want the total distance travelled after 10 seconds have passed Use the formula to work that out.

  11. welshfella
    • one year ago
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    Total distance = (n/2) * [a1 + L)

  12. welshfella
    • one year ago
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    work ouy n/2 first then a1 + L then multiply the 2 results

  13. Purplemexican
    • one year ago
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    i get 265

  14. welshfella
    • one year ago
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    that's right

  15. Purplemexican
    • one year ago
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    thank you

  16. welshfella
    • one year ago
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    yw

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