Equations of Motion by Graphical Method

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Precisely

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[@Gust52]

10:23 - To find the Displacement, first of all, we can see that the area of this region is enclosed in a Quadrilateral, so let the *Quadrilateral formed be named OABD* (Naming from *Bottom-Left* to the *Top-Left* of the Quadrilateral) . We can see that *OD || AB, but BD is NOT || OA, therefore the Quadrilateral is a Trapezium/Trapezoid.* Let *P be a pt on AB* , such that *DP is the corresponding height of AB.* Since *ar(Trapezium) = [(a+b)/2]×h* where *a & b are the || sides, and h is the perp. dist. between the || sides,* ==> *ar(OABD) = [(OD+AB)/2]×DP* ==> *ar(OABD) = {[(50+100)/2]×11} m* ==> *ar(OABD) = [(150/2) × 11] m* ==> *ar(OABD) = (75 × 11) m* ==> *ar(OABD) = 825 m* *∴ The body has been displaced by 825 m in 11 s, away from the starting point.*

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There is also a algebraic method you know or not ??

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If not first learn that

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[@gsh20230]

Sir, to derive 2nd equation with graph 24:17 Can we directly use area of trapezium formulae - > 1/2(sum of parallel sides) h I did by this, and derived it Will it be right method?

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I used to memorize but now i understand...thank you 🙏

[@AdityaHiremath-n4p]

Sir please teach this derivation using calculus method for class 11

[@MartinezTech-jw7dy]

Thanks for making me understand how to derived Equations of motion

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For everyone who is trying to learn derivations take a note and just try doing it . If u think about it u won't be able to do it

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Nicely Explained.......thanks !

[@Laureli365]

So great

[@purshotamgami2037]

After 11:03 min, you introduced 3) area = xy = vt- this is not same as (0.5*u+v)average velocity x time

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[@rajvinderkaur5956]

We can also prove the 3rd equation by 2nd equation just put the value of t and eliminate

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Ready to hit 1 million subscribers

[@KhinYadanar-l7e]

our teacher say to read the equcation but this teacher say how become the equcation by graphial method👍

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[@rayanfahad7923]

BEAUTIFUL EXPLANATION

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Sir please explain about oblique projectile motion

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[@IsraelDesta-e2d]

Thank u for best explanation

[@snowflake7624]

From velocity time graph we will get displacement or distance??

[@avigaming7723]

From velocity time graph you get acceleration By area enclosed under a graph you get distance..

[@amnajaamna]

If the acceleration is in a straight line in specified direction (meaning it's constant and not variable) then distance and displacement can be used interchangebly in a velocity-time graph.

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[@Blackkitty2023]

Good explanation Sir

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Thanks

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10th like

[@abbanmuneesa]

31:17 you did wrong 😂

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Great video. Thank you.

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Thank you.

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Thank😊

[@pandugamer8610]

Sir,I am studying in class 9. I have a doubt regarding this topic. Why do we take the initial velocity as non-zero at time 0 seconds on Velocity-Time graph to derive equations of motions. Actually,Can velocity be non-zero when time taken is 0 ? Sir,I am very confused. Sir,Please clear my doubt..🙏

[@rik_77]

VERY NICE SIR! Thanks

[@siddhardhabanka7363]

sir in distance calucation in 5 sec we are actually caluclating the area under the graph so area units are m^2 but at the end of solution we are using distance units meters how? isn't contrast the units or is there any point i miss here could you pls let me clarify my doubt pls

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10:23

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Good sir

[@rakeshmaandre4456]

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[@_sarath_official_8438]

Third derivation is quite confusion

[@abbanmuneesa]

Sir when we put vsq. At right side the u became - u

[@omxky]

Sir I am currently in 11th, your video got recommended to me while I was revising these derivations. So I wanted to say thanks because you really helped in class 9th and 10th. I was able to secure 94% in class 10. Thankyou Sir

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Congratulations 👏 on your results! Great to hear that your videos were helpful! We are uploading Class 11 videos also on our channel! So you can watch them too :)

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Nice session

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[@_sarath_official_8438]

Because, how can height be t. And parallel line would be v

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