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figure 1

As shown in the figure:

  • (+) direction is upwards
  • displacement is y
  • time=t

Let's say that $ y=30-5t^{2} $ , thus the second derivative will give acceleration as -10 which is (-) as expected. Now let's say $$ y=30 + 5e^{-t} $$ as t will increase from 0 to infinity, y will decrease from 35 to 30, thus it's falling. But second derivative, the acceleration is $ 5e^{-t} $ which is always positive for t>0. So, how can a falling object have positive acceleration when it's falling, what am I missing here?

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  • $\begingroup$ Please correct me if I am wrong but you cannot assign arbitrary equations to any "physical" system and expect them to satisfy it (the system). I believe that the second equation you have provided just does not satisfy the system. $\endgroup$
    – ZaellixA
    Commented Apr 14, 2021 at 19:03
  • $\begingroup$ Think about a simpler example, with no equations required. Suppose you drive a car along a straight road. You start from rest, accelerate up to 30 mph, then slow down and stop. Your velocity is always positive. Your acceleration is positive while you are speeding up, and negative while you are slowing down. $\endgroup$
    – alephzero
    Commented Apr 14, 2021 at 19:05
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    $\begingroup$ @ZaellixA it would be quite easy to make a physical system that behaves like the OP's equations. Think about an overdamped oscillator, for example. $\endgroup$
    – alephzero
    Commented Apr 14, 2021 at 19:07
  • $\begingroup$ Thanks for the comment @alephzero. I didn't really put any effort on checking this specific equation used by the OP so this may have led to incorrect conclusions (on my side). Thanks for the clarification and the help :). $\endgroup$
    – ZaellixA
    Commented Apr 16, 2021 at 9:05

1 Answer 1

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At $t=0$ the particle has negative velocity $v=-5$. If it were to carry on at this velocity it would sail on past the origin and head into the negative $y$ direction. It does not do this - its speed decreases as it approaches the origin and it eventually comes to a rest at $y=0$ in the limit $t\rightarrow\infty$. In order for this to happen it must be accelerating in the positive $y$ direction so that it's speed towards the negative direction decreases with time.

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  • $\begingroup$ Oh true, there is an initial velocity. Writing the equation, I've just thought it as if the object was released with a 0 initial velocity which actually isn't the case with the equation I wrote. If I wanted that, the equation should've been: 5e^(-t) + 5t + -25 .Thank you so much! $\endgroup$
    – Inpark Link
    Commented Apr 14, 2021 at 19:47
  • $\begingroup$ You're very welcome! Btw I forgot to add 30 in my answer, not that it makes any difference. $\endgroup$
    – xzd209
    Commented Apr 14, 2021 at 19:52

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