Acceleration question,

[Reiner04]

So basically this
We don't assume any acceleration happening if we have assumed that constant velocity had taken place to find K.E. or that we don't assume constant acceleration at all due to not knowing the rate of change of velocity, but only constant velocity, yet I've seen many calcs (i wouldn't doubt that there are hundreds of them), in which users has calculated acc of an object despite having assumed constant velocity to find KE or has assumed constant acc. So i wanted to ask if it's just choice of preference or we strictly do abide by it? And if so, how we are supposed to get acceleration in such cases where object went from 0 to V velocity?

 

[Reiner04]

Badump

 

[KLOL506]

@DontTalkDT

 

[Reiner04]

Disappointed badump.

 

[Antvasima]

@Executor_N0 @Spinosaurus75DinosaurFan @Mr._Bambu @Therefir @Ugarik @DMUA @Damage3245 @TheRustyOne @DemonGodMitchAubin @Jasonsith @Wokistan @Armorchompy @Migue79 @Psychomaster35 @CloverDragon03 @Dark-Carioca @AbaddonTheDisappointment @Aguywhodoesthings @Agnaa @Dalesean027 @DemiiPowa @Flashlight237 @SeijiSetto @Chariot190 @SunDaGamer

What do you think about this?

 

[SeijiSetto]

from what i can parse DT is just saying "given a distance s and a time t, assume v to work like standard velocity (s/t), but if you want acceleration for force, you assume they went from 0 to v in t seconds, i.e constant acceleration"

 

[Reiner04]

from what i can parse DT is just saying "given a distance s and a time t, assume v to work like standard velocity (s/t), but if you want acceleration for force, you assume they went from 0 to v in t seconds, i.e constant acceleration"

So incase to get acc, I'll just use constant velocity i got in first part as a final velocity for whole time period?

 

[SeijiSetto]

So incase to get acc, I'll just use constant velocity i got in first part as a final velocity for whole time period?

yeah.
say s is 10m and t is 2s. v is then 5m/s.
if you wanted acceleration, you do change in velocity over time, where v from earlier is your final velocity.
(5 - 0)/2 = 2.5m/s^2.

 

[Reiner04]

yeah.
say s is 10m and t is 2s. v is then 5m/s.
if you wanted acceleration, you do change in velocity over time, where v from earlier is your final velocity.
(5 - 0)/2 = 2.5m/s^2.

Okaish, thank you! That clears up my doubts.

 

[Dalesean027]

So this is settled then?

 

[Reiner04]

So this is settled then?

Yes, ig so.

 

[Antvasima]

I will close this thread then. Thank you to everybody who helped out here.