i guess Im randomly asking but does anybody know of a method to get back into an Instagram account? I somehow lost my password. I appreciate any help you can give me
Shouldn’t the first srcond and last part be curves? As the velocity is increasing ehich means the gradient of distance time graph should be increasing. But you drew constant gradient in all the parts
Last time you said the displacement is 400. And now you are saying that the distance is 400. So what is the difference between distance and displacement? According to you. To me they are different. Meanwhile you are the only one who give me clear answer how to translate velocity- time graph to distance-time graph. And I appreciate your work.
For the time intervals, shouldn't the lines be curves in the distance time graph? In the first 5 seconds, the object was traveling at an increasing speed and reached a max speed of 10m/s and a constant acceleration of 2m/s/s. So if we show a straight line from t (0-5) in the distance time graph, that would mean as if the object was traveling at a constant speed because of the constant gradient in the distance time graph that represents speed. So that should be a steeper curve from t(0-5). For t (5-10), the curve would become more steeper than the curve for t(0-5) and becomes a straight line (with positive gradient) for t(10-15). For t(15-20), the curve would show positive gradient with decreasing speed( curve becomes less steep and towards horizontal)
Sir the gradient of a velocity time graph represents the acceleration therefore only a line with a gradient of zero is constant velocity.. whereas on distance time graph the gradient represents velocity. And what you did is that you sketched the constant acceleration on the v-t graph as constant velocity on the d-t graph when u have drawn a the line with a constant gradient which is information. The gradient for the first 10s should be increasing, then constant from t=10 to t=15, and finally the gradient should decrease from t=15 to t=20 .
Yes you are right. Thanks for pointing that out. First two lines (red and black) should be concave up as the velocity is increasing. Third (green) should be straight as shown. Fourth (red) should be concave down as the velocity is decreasing (slope negative) Thanks Andrew
URghhh!!! i watched your video just because i wanted to know how to draw the curve when plotting a d-t graph. :/ Can u pls upload a new video correcting this major mistake ? :(
There are different levels at which students are learning the same topic. Here is the video which can help you: ruclips.net/video/MFpnow53fVs/видео.html Let me know. Thanks
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Why are you like the only person on RUclips who inputs values instead of just sketch the graph? Thank you btw
Thanks. I hope this approach is better to understand the concept. Appreciate your comments
Hey, btw the last time, should it be going down to 150 on the Y axis?
Thanks. Distance is always positive. So it always adds up. Hope that helps.
Anil Kumar what happens if your displacement is below the x axis on the velocity Tim egraph. How do you convert that onto your distance time graph
i guess Im randomly asking but does anybody know of a method to get back into an Instagram account?
I somehow lost my password. I appreciate any help you can give me
Sir, You made this way more easier than i thought. Thank you very much. Stay safe
Shouldn’t the first srcond and last part be curves? As the velocity is increasing ehich means the gradient of distance time graph should be increasing. But you drew constant gradient in all the parts
We are showing distance, which is increasing with time as you move. Hope that works. Thanks
Nice explanation! Thank you for the great video lessons.
Thanks for appreciation.
how come you didn't put any curves on the dt graph to show that it is speeding up or slowing down?
Great sir
Last time you said the displacement is 400. And now you are saying that the distance is 400.
So what is the difference between distance and displacement? According to you. To me they are different. Meanwhile you are the only one who give me clear answer how to translate velocity- time graph to distance-time graph. And I appreciate your work.
For the time intervals, shouldn't the lines be curves in the distance time graph?
In the first 5 seconds, the object was traveling at an increasing speed and reached a max speed of 10m/s and a constant acceleration of 2m/s/s. So if we show a straight line from t (0-5) in the distance time graph, that would mean as if the object was traveling at a constant speed because of the constant gradient in the distance time graph that represents speed. So that should be a steeper curve from t(0-5).
For t (5-10), the curve would become more steeper than the curve for t(0-5) and becomes a straight line (with positive gradient) for t(10-15).
For t(15-20), the curve would show positive gradient with decreasing speed( curve becomes less steep and towards horizontal)
By the way you are telling correct and sir is telling wrong
Mr. Kumar, excellent explanation, it helped a lot :)
Thank you sir. more physics tricks for NEET
Average velocity is 20m/s?
Sir the gradient of a velocity time graph represents the acceleration therefore only a line with a gradient of zero is constant velocity.. whereas on distance time graph the gradient represents velocity. And what you did is that you sketched the constant acceleration on the v-t graph as constant velocity on the d-t graph when u have drawn a the line with a constant gradient which is information. The gradient for the first 10s should be increasing, then constant from t=10 to t=15, and finally the gradient should decrease from t=15 to t=20 .
Wow Sir. This was awesome. Thank you so much sir.
Thanks for appreciation.
sir you are great!
Very informative ❤
Cool!🔥
realy helpful my confusions clear...
Thanks so much! Your video helped a lot! :)
Using trapezium formula 4 finding areas will b quick way 2 get answer.
thank you for the clear explanation! i understand now :D
Thank you sir ...
supposed to be curved?????
Yes you are right. Thanks for pointing that out.
First two lines (red and black) should be concave up as the velocity is increasing.
Third (green) should be straight as shown.
Fourth (red) should be concave down as the velocity is decreasing (slope negative)
Thanks Andrew
URghhh!!! i watched your video just because i wanted to know how to draw the curve when plotting a d-t graph. :/ Can u pls upload a new video correcting this major mistake ? :(
@@ruvindrigunawardena3369 sameee.. Did u figure it out?
Thnku sir
where
very helpful thanks
Thanks for appreciation
for which standard it is
9 or 10 or 11
I don't agree with this teaching, I think it doesn't fully explains the cure of the graph, which are changing at each inflection points.
There are different levels at which students are learning the same topic. Here is the video which can help you: ruclips.net/video/MFpnow53fVs/видео.html
Let me know.
Thanks
is the average velocity 15?
average velocity=total distance/total time = 400/20 = 20 s
Thanks
Can I get get your introduction including education