Distance traveled at end of second interval: 27m

[Ki-nana18]

If an object falls five meters during the first interval of time, what is the total distance fallen at the end of the second interval of time? (Galileo's Theory of falling bodies)

I know that at successive intervals of time the distance fallen is proportional to the odd numbers. So I suspect it to be 12, but apparently I'm wrong.

 

[cepheid]

The distance traveled by a body moving under constant acceleration is given by a well-known equation that is derived from Newton's laws. There is no need to guess! I would look into this equation...

Further comments

It doesn't really make much sense to me to break up the problem into discrete steps like this. Also, "proportional to the odd numbers" doesn't make much sense to me. Even more strangely, your guess, 12, is an even number.

EDIT: On second thought, once you know the formula, then you can break it up into steps (time intervals Δt) and you will see that the ratio of the *total* (cumulative) distance covered to the distance covered during the first time interval does depend upon the number (n) of intervals Δt that have occurred in a very specific way.

 

[turin]

Galileo observed objects rolling down inclined planes. Using a pendulum as a timer, he observed that the distance of travel increased for every period of the pendulum, by consecutive odd number multiples of the distance traveled in the first interval. The kinematic equation for constant acceleration can be shown to agree with this observation, and it suggests a particular relationship between distance and time. (Calculate ∑{(2n-1)^2} where n is the interval number from n=1 to N, and then observe the dependence of N on n. The distance traveled is Nd, where d is the distance traveled in the first interval.) This is one way that Galileo determined the constant universal gravitational acceleration.

See, for example:
http://galileo.rice.edu/lib/student_work/experiment95/inclined_plane.html

Ki-nana:
It looks like you added 5m+7m=12m. That would be valid if the object traveled 5m in the third interval (and then 7m in the fourth interval). (5=2(3)-1, 7=2(4)-1) However, it travels 5m in the first interval, so the appropriate odd numbers are 1=2(1)-1 and 3=2(2)-1. Then, just multiply by the distance in the first interval.