I have two questions: 1. In a displacement-time graph, an increasing gradient or a curved line upwards means the velocity is increasing rapidly and we would use the tangent line to calculate the gradient at any given point, right? 2. How would you calculate, in a velocity-time graph the area(displacement) under a curved line?
Ans.1- Yes. An increasing gradient or a curved line upwards means the velocity is increasing rapidly . The slope of tangent line at a point on the curve, tells the instantaneous velocity at that point. If slope makes angle θ with the time axis, tanθ will be instantaneous velocity. Ans.2.- To calculate area which is equal to displacement, in a velocity-time graph under a curved line, we will take help of integrals calculus. We will divide the whole area into infinite numbers of strips i.e. No. of Strips N→∞.If there is expression which shows relation between v and t, Area A= ∫ v(t)dt. For details keep watching physicsnama next coming video.
Thanks dale for you keen narrative skills which go smoothly into my mind and joe the script was was amazing and beautiful with simple and good vocabulary, well that is what peopl6are searching for . Keep it up 😀😀
i just found out about this channel and i think u guys are really cool (the animations are great and really help to us to understand the concept more practically) but it would be better if the speaker (the one explaining the concept) had a bit more enthusiasm and energy in his voice
Some hard trap playing in the background
Everyone liked that
ikr
kinda sounded like ear rape
Teachers should teach like this more often! This was a really great explanation
Glad you think so!
Hello tae
I really admire your animations. I wish I could replicate that to some extent for my students. Keep up the excellent work!
Thank you very much!
this channel should have more viewers.
Thanks! Appreciate it!
I have two questions:
1. In a displacement-time graph, an increasing gradient or a curved line upwards means the velocity is increasing rapidly and we would use the tangent line to calculate the gradient at any given point, right?
2. How would you calculate, in a velocity-time graph the area(displacement) under a curved line?
Ans.1- Yes. An increasing gradient or a curved line upwards means the velocity is increasing rapidly . The slope of tangent line at a point on the curve, tells the instantaneous velocity at that point. If slope makes angle θ with the time axis, tanθ will be instantaneous velocity.
Ans.2.- To calculate area which is equal to displacement, in a velocity-time graph under a curved line, we will take help of integrals calculus. We will divide the whole area into infinite numbers of strips i.e. No. of Strips N→∞.If there is expression which shows relation between v and t, Area A= ∫ v(t)dt. For details keep watching physicsnama next coming video.
I really love Fuse school its very education keep up the good work
Thank you :) much appreciated!
@@fuseschool is one good educational
Thanks dale for you keen narrative skills which go smoothly into my mind and joe the script was was amazing and beautiful with simple and good vocabulary, well that is what peopl6are searching for . Keep it up 😀😀
Wow, thank you so much! 💜
Great explanation!
Thank you!
understood perfectly great explanation
Great to hear!
THANK YOU SO MUCH THIS HELPED ME AT THE VERGE OF MENTAL BREAKDOWN
Glad we could help. Take care of yourself!
They are stunning videos. Thank You so much
Thank you too! Glad you like them!
It is very helpful video
Glad it was helpful!
Amazing video... Thanks for your help..
Thanks! Glad you liked it 🙂
Nice🤯😊
Thanks 😋
Very clear. More pls
Thank you 😉
Thanks so much
Really great
Most welcome!
@0:16 Street fighter "Perfect!" Sound effect
Beautiful!
Thank you!
i just found out about this channel and i think u guys are really cool (the animations are great and really help to us to understand the concept more practically) but it would be better if the speaker (the one explaining the concept) had a bit more enthusiasm and energy in his voice
Thank you! And thanks for your feedback!
Great video really helpful
Glad you think so!
Can u explain this more detailed in another video. I didn't understand anything.
Lol same
Please provide me link of distance time graph video.
th-cam.com/video/lrHAnr9cYdQ/w-d-xo.html
point be is constant velocity and therefore no acceleration
FuseSchool can you clarify as I agree with the comment, a flat line indicates 0m/s2 due to no change in velocity = 0m/s divided by time = 0m/s2
@@mramod6307 That's right, no change in velocity means acceleration is zero. Your calculation is correct.
Song name?
Please, check the end credits 😉
@@fuseschool I cant find the song
If you have access to TH-cam Audio Library, search for "CxDy - Turbo"
@@fuseschool thanks
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Ti teg tnod I
Lol teacher
Simply waste