Question 161461
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I don't have any idea how to do these word problems. I've been trying but I just don't get it. Can you please help me?
I can never understand word problems.

1) The distance an object falls is directly proportional to the square of the time it has been falling. After 6 seconds
it has fallen 1296 feet. How longwill it take to fall 2304 feet?

2) x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled?

If you can help I'd really appreciate it. Thank you so much! I could not do this without you.


<font size = 4><font color = blue><b>1) The distance an object falls is directly proportional to the square of the time it has been falling.
   After 6 seconds, it has fallen 1296 feet. How long will it take to fall 2304 feet?</font></font></b>

      D = {{{kT^2}}}
  1,296 = {{{k(6^2)}}} ----- Substituting 1,296 for D (distance), and 6 for T (time) 
  1,296 = 36k
{{{"1,296"/36}}} = k
     36 = k


       D = {{{kT^2}}}
   2,304 = {{{36T^2}}} --- Substituting 2,304 for D (distance), and 36 for k
 {{{"2,304"/36}}} = {{{T^2}}}
      64 = {{{T^2}}}

<font color = red><font size = 4><b>Time taken by object to fall 2,304 feet</font></font></b>, or {{{T = sqrt(64)}}} = <font color = red><font size = 4><b>8 secs </font></font></b>
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<font size = 4><font color = blue><b>2) x varies directly as the square of s and inversely as t. How does x change when s is doubled?
   When both s and t are doubled?</font></font></b>

<font size = 4><font color = red><b>2a) How does x change when s is doubled? </font></font></b>


With this being DIRECT, and INDIRECT/INVERSE VARIATION, and with k being the CONSTANT of PROPORTIONALITY, we get the following equation: 

x = {{{k(s^2/t)}}} 
x = {{{k(((2s)^2)/t)}}} ---- Doubling s, or replacing s with 2s
x = {{{k((4s^2)/t)}}}  
x = {{{highlight(4)k((s^2)/t)}}} ---- Equation, after s is DOUBLED
x =    {{{k((s^2)/t)}}} ---- ORIGINAL equation

Upon comparing the 2 equations above, it’s clearly seen that, <font size = 4><font color = red><b>when s is DOUBLED, x is QUADRUPLED</font></font></b>.
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<font size = 4><font color = red><b>2b) How does x change when both s and t are doubled?</font></font></b>


With this being DIRECT, and INDIRECT/INVERSE VARIATION, and with k being the CONSTANT of PROPORTIONALITY, we get the following equation: 

x = {{{k(s^2/t)}}} 
x = {{{k(((2s)^2)/(2t))}}} ---- Doubling “s” and “t”
x = {{{k((4s^2)/(2t))}}}
x = {{{k((2cross(4)s^2)/(cross(2)t))}}}  
x = {{{k((2s^2)/t)}}}  
x = {{{highlight(2)k((s^2)/t)}}} -- Equation, after “s” and “t” are DOUBLED
x =    {{{k(s^2/t)}}} ---- ORIGINAL equation

Upon comparing the 2 equations above, it’s clearly seen that, <font size = 4><font color = red><b>when s and t are DOUBLED, x is DOUBLED also</font></font></b>.</pre>