1
$\begingroup$

I've looked up and understand the reason behind the number of the acceleration of gravity $(9.8)$. But whenever people describe it they say it's $9.8 \text{m/s}^2$.

I don't understand why it's inverse time squared, why isn't it just a number like $9.8 \text{m/s}$?

$\endgroup$
  • $\begingroup$ By the way, it's not $(\frac m s )^2$, it's $\frac m {s^2}$; which may be part of the confusion by the sounds of it. $\endgroup$ – JMac Jul 4 '17 at 23:08
  • $\begingroup$ Acceleration is the rate of change of speed, that is how fast does the speed change per unit of time. In the case of gravity it is changing at 9.8 meters per second every second. As an example, initially at speed = 0 m/s, at t=1 second it is 9.8 m/s, at t=2 seconds it is 2x9.8 m/s=19.6 m/s etc. $\endgroup$ – nluigi Jul 5 '17 at 8:11
3
$\begingroup$

As any ordinary acceleration, the quantity $9.8\, \mathrm{m/s^2}$ means that the velocity increases by $9.8$ meters per second, each second. Hence $9.8$ meters per second, per second.

$\endgroup$
  • $\begingroup$ I feel silly for not getting it, but it's still not clicking. If my velocity started at 0 and I was falling for 4 seconds then i'd be moving at 9.8 * 4 = 39.2 (focusing just on the acceleration and ignoring mass/air resistance etc). I just don't get where the squared part factors into any of this, or is what I wrote wrong? Or is the squared part only relevant to the units of measurement? $\endgroup$ – user1157885 Jul 5 '17 at 6:52
  • 2
    $\begingroup$ @user1157885 Hint: what units does the 9.8 have, if 4 is in seconds and 39.2 is in m/s? $\endgroup$ – probably_someone Jul 5 '17 at 7:02
1
$\begingroup$

Well, what is the definition of acceleration(due to gravity or otherwise)? It is

$$ \frac{\Delta v }{ \Delta t},$$

or, in words: it is the instantaneous change in velocity from one point to another divided by the associated time interval.

The unit for velocity is $m/s$. The unit for time is $s$. Hence, the units for acceleration are $m/s/s = m/s^2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.