Recently i came accross a problem that said

An object is dropped straight down from helicopter the object falls faster and faster but its acceleration decreases over time becoz of air resistance. the acceleration is recorded every second after the drop for 5 second as below,

at t=0 secs, a=32ft/sec^2;

at t=1 sec, a= 19.41ft/sec^2;

at t=2 sec, a=11.77ft/sec^2;

at t=3 sec, a= 7.14ft/ sec^2;

at t= 4 secs, a=4.33ft/sec^2;

at t= 5 sec, a=2.63 ft/sec^2;

when i calculated the velocities they were as follows,

at t=0 secs, v=0;

t=1 sec,v= 32ft/sec;

at t=2, v=51.41;

t=3, v=63.18;

t=4, v=70.32;

t=5 v=74.65.

here as you can see v2 minus v1 gives the acceleration, but here it isnt average acceleration,they mentioned that acceleration is recorded at every second. so they mean its instantaneous acceleration, and if it is instantaneous acceleration then how come the change is happening over an interval of 1sec?? can anyone explain without the use of kinematical equations? would be grateful, thank you! read below if you didnt get wat m saying

i found this in a calculus book,there was this in calculus book,thomas calculus,i was studying finite sum approximation and had to find upper and lower estimate, out of curiosity i started to think practically,forget about right or wrong. hey wat if incase u come accross such situation, i mean just read the last few lines of my question thats wat i want to know.let me explain.

wat m trying to say here is u ll observe that they said the acceleration is recorded every second, meaning instantaneous acceleration at every second is given,

then just look at the data at t=1 sec, a= 19.41ft/sec^2; i.e its the instantaneous acceleration at t=1 sec, are u clear till here??

now see the velocity at t=1 sec its v= 32ft/sec;

now look at the velocity at t=2 sec its 51.41ft/s

i.e v2 -v1=19.41ft/s^2 i.e acceleration at t=1 sec i.e wat they call instantaneous acceleration at t=1 sec.(like they said acceleration is recorded at every second)

how come the velocity at t=2 sec be 51.41ft/s. i mean 19.41ft/sec^2 is instantaneous acceleration so how come the change happened over an interval of 1 sec.

think of it as a case. i hope you get wat m saying.

  • $\begingroup$ It is unclear to me what is the question $\endgroup$
    – user65081
    Jan 26, 2021 at 22:04

2 Answers 2


We model nature continuously, with differentiable derivatives; a position function has a nice velocity function, a velocity function has a nice acceleration function, and an acceleration function has a nice jerk function. This is how Classical Mechanics is modelled.

However we cannot observe continuous things in the strict sense, because all of our observations are 'snapshots' per se. So when we calculate any derivative of a real function; we are actually computing very small secant line since our tools are not able to do infinitesimal limits.

Think about how we even measure speed. Is there really a way to calculate 'instantaneous velocity'? It's a subtle question but if you think about you will realise no; we can only measure the change in position over the change in time, and make those changes in time smaller and smaller, but never actually 0.

So try and think about what the question is implying with instantaneous acceleration; since indeed if we know it at 0 seconds and 1 second, we have no idea if it shot up to something wild at 0.5 seconds.

There is a lot of work done adjacent to this question involved with Numerical Modelling, essentially using discrete systems to simulate Analog Phenomena, if you are interested more in the question of 'instantaneous'

  • 1
    $\begingroup$ That's a decent answer. +1 from me. $\endgroup$
    – Gert
    Jan 26, 2021 at 19:06
  • 1
    $\begingroup$ To add some math to this good answer, the acceleration is $$a=\dfrac{dv}{dt}$$ but how we calculate it is: $$a=\dfrac{\delta v}{\delta t}$$ where the $\delta$ denotes a very small but finite change. $\endgroup$
    – user256872
    Jan 26, 2021 at 22:06

If at time 0 acceleration is 32, and at time 1 acceleration is 19.41, then the average acceleration in this time span is (32+19.41)/2 = 25.705.

Apply this average acceleration as a constant value starting from speed 0 to get the speed at time 1 as

speed(t=1) = 0 + 25.705 * (1 second) = 25.705

If you proceed for the remaining time step you can calculate the (approximate) velocities at each time step.

speed(i) = speed(i-1) + ( accel(i)+accel(i-1) )/2 * (time step)
time [s] acceleration [ft/s^2] speed [ft/s]
0 32 0
1 19.41 25.705
2 11.77 41.295
3 7.14 50.75
4 4.33 56.485
5 2.63 59.965

Then you fit a curve through the acceleration vs. speed data to find what kind of air resistance you have

data fit

To me it seems you have linear air resistance with the following model

a = g - b*v

use the data above to estimate the parameters g and v. Next, predict the acceleration at various points to see how well it matches the given data.

The fact that the model fits the data is confirmation of the assumption of taking the average acceleration for each time step is sufficient to this model.

So the data answers your question by saying that although acceleration varies continuously use constant acceleration for small enough intervals of time gives us an appropriate mathematical model.


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