Acceleration due to gravity?

Question

Is measured at 9.8 meters per second squared. Why is second squared? Does a ball dropped in a vacuum square it's speed every second that it is falling?
What the heck does "per second per second" mean?

Answer 1

You should probably skip what I wrote and go straight to the link http://whatis.techtarget.com/definition/0,,sid9_gci529957,00.html :-)

You measure the distance fallen in meters. Let us say that this distance as a function of time t in seconds since the ball was dropped in a vacuum is s(t) meters.

The "instantaneous" downward velocity of the ball at time t is just the derivative of s(t) with respect to t ... for example in the h seconds between time t = k and time t = k + h, the ball drops an extra s(k + h) - s(k) meters ... so the average velocity between times t = k and t = k + h is (s(k + h) - s(k)) / h, and the limit as h -> 0 of (s(k + h) - s(k)) / h is the velocity at time t = k and is the derivative of s(t) at t = k, and you can see that it has units meters per second because s(k+h) - s(k) is in meters and h is in seconds.

So distance is meters and velocity is meters per second, with v(t) = s'(t) the derivative of s with respect to t.

Now acceleration is defined as the rate of change with repsect to time of velocity. You will have a(t) = v'(t) = s''(t), by the same reasoning as before. The velocity at time t = k is v(k) = s'(k) and the velocity at time t = k + h seconds is v(k + h) = s'(k + h). The average acceleration between times t = k and t = k + h is then (v(k + h) - v(k)) / h in units of (meters per second) per second, or meters per second squared. The instantaneous acceleration at time t = k is the limit of that as h -> 0 and is v'(t) = s''(t) at t = k.

Falling in a vacuum at constant acceleration g = 9.8 meters per second squared means you have

acceleration = constant g = 9.8 meters per second squared

increased downward velocity after falling t seconds = g t = (9.8 meters per second squared) time (t seconds) = 9.8 t meters per second.

distance fallen (if dropped with initial velocity zero) after falling t seconds is (1/2) g t^2 = (1/2) (9.8 meters per second squared) (t seconds)^2 = (1/2) g t^2 meters.

The rate of change of acceleration would be meters per second cubed.

Answer 2

Its velocty increases at a rate of 9.8 meters per second every second, thus "meters per second PER second". It's velocity is not squared every second.

The "seconds squared" part is an imaginary unit used by convention. No one has ever observed a squared second; there are no devices that measure things in units of "seconds squared". It merely expresses that a rate of change (of position) over time is undergoing change over time.

Answer 3

Acceleration due to gravity of per second every second means gaining additional speed of 9.8 meters every second.

Answer 4

yes it does

the second squared term is just a math way of saying it

it makes more sense to express it as meters per second per second

in other words, every second that the object falls, its velocity is getting 9.8 meters per second faster

after 1 second, the object is going at a speed of 9.8 meters per second

after another second, it is going 19.6 meters per second

see, 9.8 meters per second velocity per second of falling

Answer 5

The acceleration due to gravity of a ball in a vacuum would be calculated the same way because gravity still has an effect in a vacuum.

Answer 6

it means "in every second, the speed of the fallling object increases by 9.8 meters per second .... "

(9.8 m/sec) /sec = 9.8 m/sec^2

does this make sense?

Answer 7

accel is velocity per sec

velocity is meters per sec

thus per sec per sec