I am not a physics professional. I just have found a casual interest in physics. I have a question about gravity. Gravity accelerates objects at 9.8 meters per second, per second. I will round this off to 10 meters per second per second. Does this mean that for every second, the rate of acceleration increases? In other words, after the 2nd second, the rate of acceleration is 20 meters per second? After the third second, the rate of acceleration is 30 meters per second? Is the RATE of acceleration increasing by 10 meters per second for every second so that after 20 seconds the rate of acceleration would be 200 meters per second?
No, it's the speed that is increasing, not the acceleration. Speed is in meters per second and that speed increases as an object falls. The acceleration is the rate of change of that speed, in meters per second per second. That acceleration is a constant.
Ok. So the acceleration remains at 10 meters per second then? If that's true, why is it said that the rate of acceleration due to gravity is 10 meters per second PER SECOND? Why is that 2nd "per second" there? Why don't they just say the rate of acceleration is 10 meters per second? Why 10 meter per second PER SECOND?
Meters per second is the units for speed, not acceleration. The acceleration is 10 m/s per second. That means every second the speed increases by 10 m/s.
Thanks for clearing that up. I had some confusion about whether it was just the speed that was increasing, or both the speed and the rate of the acceleration. Seems like you've cleared that up for me. Is there a math equation that describes the rate of acceleration due to gravity? My father told me he had seen an equation with t squared in it. Are you familiar with any such equation?
Good! Here are some equations that describe falling motion under gravity. a = g = 9.8 m/s/s. (This just says that the acceleration is a constant 'g', which is 9.8 m/s/s.) v = a*t = g*t (This gives the speed of a falling body at any time after it's dropped. So, after 1.5 seconds, v = 9.8*1.5 = 14.7 m/s.) s = 1/2*a*t^2 = 1/2*g*t^2 (This is the equation you were looking for. It gives the distance fallen, s, as a function of time. So, after 1.5 seconds, the object falls a distance of s = .5*9.8*1.5^2 = 11 m.)
That sounds like the equation for the distance an object falls under the influence of gravity. If you drop an object from rest, after t seconds it has fallen the distance $$s = \frac{1}{2}gt^2$$ (Added: Doc Al beat me to it while I was typing...)