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?
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
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
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?
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.
![[TH] 2024 PMSL SEA W3D1 | Fall | สัปดาห์ตัดสิน ทีมไปไฟนอล](http://i.ytimg.com/vi/ZBU3GyzqErI/mqdefault.jpg)
Very well explained, useful animations. Thanks a lot.
Lot of love from 🇵🇰
A distinctive, beautiful and full explanation of all the details of the lesson
I benefited a lot
Thank you very much for this video
Superbly explained.. thanku
u save my life. Danke sehr 👍🏻
helped me so much, thanks
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?
same
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
where did you get this v-t graph which we can show terminal velocity..thanks
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! ;)
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
How do we know when to take 9.81 ms-² as positive and when as negative?
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?
Why is drag downwards positive in your table . Thankyou :)
Cuz in this system we've defined all vectors pointing downwards to be (+)! You can choose ;)
If the net force is upwards when the parachute is opened then why is the object still coming down
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.
I wanna skydive
13:43