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Sir please solve this I have exam on Sunday 18. a) Acceleration is the time rate at which velocity of a body changes. Show that for a car, moving with constant acceleration for a period of time, the distance travelled in the second half is three times of that in the first half. b) The acceleration caused by the earth’s attraction towards its centre is known as acceleration due to gravity. It is always directed towards the centre of earth. Show that in the absence of air resistance, for a ball thrown upwards, the time of ascent is equal to the time of descent. c) Shown that when air resistance is also taken into account, the time of descent is greater than the time of ascent.

Sir please solve this I have exam on Sunday A truck is moving forward at a constant speed of 20 m/s. The driver sees a car in front of him at a distance of 110 meters stopping suddenly. After a reaction time he applies the brakes which give the truck an acceleration of -3 m/s2 . What will be the maximum allowable reaction time to avoid a collision and what distance the truck would have moved before the brakes take hold?

Sir since it depends on initial and final position . But here initial points are different in case 1(A to B) then case 2(C to B) and 3(C to B) . But still the work done is same in all the three cases . Please explain why 18:00

Amazing class

The way u make much complicated questions easy for us is commendable.Keep uploading ❤

Bhai Dil se keh raha hu jo kaam doubnut ke jhatu teacher nahi kar paye wo aapne kar diya

Thnx sir very much ❤❤

Wow sir aapke samjhaane ka bhaut aacha h ❤

Nice explanation .. though I don't know hindi.lam able to Understand what actually he is saying 👏👏

Thanks

Thanks sir!!!❤

Sirr aapki explanation just likeee wowwwwww 🤩🥳

Thnk you thank you so much sirrrr 😍 your explanation is brilliant

Sir upload all chapters of class 11 physics question

Sir please cover remaining chapters of 12 th syllabus...can't find good solution of intex questions on RUclips u r the only one who explains very well

00:03 Understanding the properties of a uniform disk with a hole 00:44 The center of the hole is located at a distance of r/2 from the center of the original disk. 01:27 Finding the center of mass after removing a portion from the disk 02:05 Finding the center of mass from origin point 02:46 Important to consider mass distribution when using formulas 03:27 Finding the relationship between m1 and x1 coordinates on a disk 04:06 Deriving the value of x2 from the center of mass of the m2 mass 04:45 Uniform disc has constant surface mass density 05:24 Finding the total mass and surface area of the original disk 06:07 Calculating the area of the removed portion 07:05 Calculate resulting mass after removing a portion 08:01 Finding the center of mass

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Thank you sir❤😊

amazing explanation sir thank you so much 😊

Sir you make physic more interesting for me thank u sir

Thank you So much sirr ❤❤❤❤

Thanks sir ❤❤

12-(-12) hona tha fir answer sahi hota I guess 😅

Kh😂

Really your hardwork is commendable!!!! Thanks a lot for solving this question in detail ✨✨

Top class explained

Thank u sir

👍

Nice explanation sir❤

Gajab ka samhghate.ho.sir😊😊

Sir is efficiency is cut from cbse

It's air not ear😢

Thanks for this good explanation

Impulse +(positive)hone ka madlab keya hota ha sir

🎉🎉 nice explanation

Ya 9.8-g kasa aya batao to

Thanks sir🎉❤

Thank you sir

you really helped me a lot!!!

Thanks you

Sir you nailed it ❤

Very hardworking man

Amazing explanation🎉

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Thank you sir nice explanation

Hii sir chapter explain krne vale vdo upload nhi honge ?

Thankyou sir love it ❤

Thank you sir for taking the efforts ...

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Its very helpful and explanation was very nice...thank you so much sir.👏👏