3.2b Terminal Velocity Graphs of Falling Object | AS Dynamics | Cambridge A Level Physics

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ETphysics

ETphysics

Күн бұрын

Пікірлер: 21
@physicsmadeeasy99
@physicsmadeeasy99 3 жыл бұрын
Very well explained, useful animations. Thanks a lot. Lot of love from 🇵🇰
@مازنسعيدالهادى
@مازنسعيدالهادى 2 жыл бұрын
A distinctive, beautiful and full explanation of all the details of the lesson I benefited a lot Thank you very much for this video
@XunaibSamar
@XunaibSamar 2 ай бұрын
THANK U so much 💕
@zafarasim3617
@zafarasim3617 3 ай бұрын
Very well explained with the help an excellent animation and step by step tabulation. It would have been even better if a point was added in the v-t graph at a lower bend, where the velocity and the drag force reduces gradually and then goes on constant as 2nd terminal velocity.
@sricharanreddy784
@sricharanreddy784 Жыл бұрын
Superbly explained.. thanku
@ytgaming_malaysia8411
@ytgaming_malaysia8411 Жыл бұрын
u save my life. Danke sehr 👍🏻
@ronnierustia6879
@ronnierustia6879 3 жыл бұрын
where did you get this v-t graph which we can show terminal velocity..thanks
@ETphysics
@ETphysics 3 жыл бұрын
We don't need to know the exact math equation for the v-t graph, but if you Google around a bit you'll see that it's actually an exponential graph! ;)
@andreayeo3262
@andreayeo3262 2 жыл бұрын
Teacher, I do not quite understand why the acceleration is decreasing at a decreasing rate? are we able to figure out that the decrease in acceleration is at a decreasing rate through calculation or is this based on concept? And if so, what is the idea behind that?
@muhammadyaameenjussub4670
@muhammadyaameenjussub4670 2 жыл бұрын
same
@raminkhan1151
@raminkhan1151 2 жыл бұрын
if u check 2.1b from the kinematics playlist you'll see that shape of the line determines whether its at constant, dec or inc rate tho its for velocity time graphs in dat video but same concept applies
@freestockvideos9212
@freestockvideos9212 2 жыл бұрын
helped me so much, thanks
@piumiperera7230
@piumiperera7230 3 жыл бұрын
teacher in the last graph of acceleration the side changed and shouldnt it approach again to zero in an decreasing rate ...rather than an increasing rate
@Imrankhan-g4h6n
@Imrankhan-g4h6n Жыл бұрын
Why is drag downwards positive in your table . Thankyou :)
@ETphysics
@ETphysics Жыл бұрын
Cuz in this system we've defined all vectors pointing downwards to be (+)! You can choose ;)
@Syeda_Fizzah_Azeem
@Syeda_Fizzah_Azeem 8 ай бұрын
How do we know when to take 9.81 ms-² as positive and when as negative?
@Syeda_Fizzah_Azeem
@Syeda_Fizzah_Azeem 8 ай бұрын
If 9.81 is taken as value of GRAVITY then it will always be negative....but if it is taken as value of regular acceleration that decreases when net force decreases, we can choose the signs ourselves?
@Imrankhan-g4h6n
@Imrankhan-g4h6n Жыл бұрын
If the net force is upwards when the parachute is opened then why is the object still coming down
@fuqqahasif7436
@fuqqahasif7436 9 ай бұрын
The net force upwards is causing acceleration in the upwards direction. Acceleration is the rate of change of velocity, so after the parachute is deployed the velocity downwards starts to decrease or another way to think (the velocity upwards starts to increase). It does not mean that the person will immediately start to rise above now. Because in order for that to happen the velocity downwards should become 0. However, that does not happen because after the parachute is deployed, your downwards velocity starts to decrease which also simultaneously decreases the air resistance as your drag force is directly proportional to velocity. And hence acceleration becomes zero again as air resistance decreases till it equals weight, reaching a new lower terminal velocity.
@amazkhawar3657
@amazkhawar3657 Жыл бұрын
13:43
@bekind6869
@bekind6869 2 жыл бұрын
I wanna skydive
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