Equations of Motion by Graphical Method 

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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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Im in year 9 now and this new topic was really confusing since the teacher didn’t even explain. He just showed the formula so I’m happy I watched this video and learnt where these formulas are coming from

Excellent demonstration

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BEAUTIFUL EXPLANATION

Thank u for the clear explanation sir😊

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Nicely Explained.......thanks !

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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Thanks for making me understand how to derived Equations of motion

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Sir please explain about oblique projectile motion

So great

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Please I need explanation on the relationship between temperature and humidity

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Thank you.

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Thank you sir for work done

Sir please explain ohms law please

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

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

Very nice explanation

VERY NICE SIR! Thanks

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

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i loved this class I want more like this

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

Superb sir

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..🙏

Thanks sir for help

Thank you so much Sir

Sir please teach this derivation using calculus method for class 11

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Good sir

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Good explanation Sir

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

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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sir we can uderstand lesson easily

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

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Tomorrow is my exam I was totally screwed until I saw this video

Kinda confused; but how is 30 the maximum velocity....I thought it should be 40 cus it's the highest number.

Can you help me

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

10th like

999k subscribers

please can I get the video before this

Third derivation is quite confusion

Hello sir, Can you please start a class on organic chemistry

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Thank you sir 😊 🙋‍♀️

10:23

Sir please explain in hindi 🙏

fr

31:17 you did wrong 😂

ഹല്ലോ മച്ചാരാതെ ചലന സമവാക്യങ്ങൾ പഠിച്ചു കഴിഞ്ഞോ? എന്ത് എളുപ്പമാണ് ഈ സമവാക്യങ്ങൾ പഠിക്കാൻ. അതിൻ്റെ നിർവചനവും എളുപ്പമാണ്.🔥🔥🔥

Sir, why don't we take value of u from zero.