DIRECT VARIATION
Problem:
» An egg is dropped from the roof of a building. The distance it falls varies directly with the square of time it falls. If it takes 0.5 second for the egg to fall 8 feet, how high must the building be if it took 6 seconds for the egg to hit the ground.
Answer:
- [tex] \color{hotpink} \bold{1,152 \: \: feet} \\ [/tex]
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Step-by-step explanation:
Based on the problem, it takes 0.5 second for the egg to fall 8 feet.
Since the distance varies directly as the square of time, then the equation of variation is in the form of d=kt²
where,
- d is the distance
- t is the square of time, and;
- k is the constant of variation
[tex] \bold{Solution: } \: \color{grey} \tt(Constant \: of \: Variation)[/tex]
- Write the equation of variation first and substitute the given values of distance.
...the time has to be multiplied by itself (squared).
- [tex] \: \: \tt \: d=kt² \\ \tt \: \: \: \: \: \: \: \: \: \: \: 8 = k(0.5)² \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: 8=k(0.25) \\ \tt \: \: \: \: \: \: \: k = \frac{8}{0.25} \\ \tt \: k = \blue{32}[/tex]
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[tex] \bold{Solution:} \: \color{grey} \tt (Unknown \: Value) [/tex]
Since the constant of variation (k) is 32 and the equation of variation is d=32t². Then, we can now find how high must the building be, given that time is 6 seconds.
Again, to find the unknown value. We just need to substitute all the givens to the equation of variation.
Given time has to be multiplied by itself (squared) so,
- [tex] \: \: \: \: \: \: \tt \: d=32t² \\ \tt \: \: \: \: \: \: \: \: \: \: \: d= 32(6)² \\ \tt \: \: \: \: \: \: \: \: \: \: \: d = 32(36) \\ \tt \: \: \: \: \: \: \: \: \: \: d = \underline{ \boxed{ \blue{ \tt {1152}}}}[/tex]
[tex] \therefore[/tex] The building must 1,152 feet.
Hence, the distance is 1,152 ft.
