SOLUTION: In the absence of friction, a freely falling body will fall about 16 feet the first second, 48 feet the next second, 80 feed the third second, 112 feet the fourth second, and so on

Algebra ->  Sequences-and-series -> SOLUTION: In the absence of friction, a freely falling body will fall about 16 feet the first second, 48 feet the next second, 80 feed the third second, 112 feet the fourth second, and so on      Log On


   

Question 1101505: In the absence of friction, a freely falling body will fall about 16 feet the first second, 48 feet the next second, 80 feed the third second, 112 feet the fourth second, and so on. How far has it fallen during:
a. the seventh second?
b. the nth second?
I am incredibly confused on how to write this problem out. I know that speed times the time equals distance, but this is a sequence not a linear equation, which is a harder concept for me to grasp so if you could show how to do this problem so that I can write it out and figure it out myself that would be wonderful.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The body is not falling at a constant rate; gravity is making it speed up. So the distance it falls in t seconds is not a linear function.

But the RATE at which the speed is INCREASING IS constant. So the formula for how far it falls DURING THE n-th SECOND is a linear function.

And you can see that from the given information. If we call the distance the body falls in the n-th second d(n), then
d(1) = 16
d(2) = 48
d(3) = 80
d(4) = 112
...

The function value is increasing by 32 each second. Then seeing that d(1) is 16, the function for the distance it falls during the n-th second is
d(n) = 32n-16.

In the 7th second, the distance it will fall is 32%287%29-16%29+=+208

We could verify that by continuing the pattern seen in the first 4 seconds:
d(5) = 144
d(6) = 176
d(7) = 208