Equations of Motion (Physics) 

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1. v = u + at = 0 + 2*5 = 10 m/s 2. v2 = u2 + 2as --> 0 = 400 + 20a a = -400/20 = -20 m/s2 3. t = u/g = 1 s (Single sided journey) Total time = 2t = 2 s s = ut + at2/2 = 10 - 5 = 5 m

ANSWER OF THE FIRST QUESTION EXPLANATION : 1. u {initial velocity} = 0 a {acceleration} = 2 m/s square t {time} = 5 s v {final velocity} = ? So here we have to find the final velocity. So we take the first equation. v = u + at = 0 + 2 x 5 = 10 m/s So the final velocity is 10 m/s So the speed or velocity of the car after 5 s is 10 m/s Please like if helpful...

my understanding just went from not having a clue what any of this meant, to being able to get it right most of the time. Great video.

Number 2 Using V2=U2+2as Convert 72km/h to m/s =72x1000/60x60= 20m/s V=0 U=20m/s S=10 a=? 0^2=(20)^2+2ax10 0=400+20a 20a=-400 a=-400/20 a=-20m/s^2 I hope Im able to help those that were so confused at first like me

Q1 - According to first equation of motion v = u + at Here , v = ? u = 0 m/s a = -2m/s square t = 5 s Put the value in equations v = 0 m/s + (2 m/s square * 5s) v = 0 m/s + 10 m/s Hence speed of the car is 10 m/s. Q2 - According to third equation of motion 2as = v square - u square Here , s = 10 m v = 0 m/s u = 72 km/h or 20 m/s Put the values in the equation 2*a*10 m = 0 square - 20 square 20 m * a = 0 - 400 m/s square a = -400 m/s square / 20 m a = -20 m/s square Q3 - According to the third equation of motion 2as = v square - u square Here , a = (-10 m/s square) because the acceleration due to gravity is in upward direction v = 0 m/s u = 10 m/s Put the values in the equation 2 * (-10 m/s square) * s = 0 square - 10 square -20 m/s square * s = -100 m/s s = -100 m/s / - 20 m/s square s = 5 m 2. According to the first equation of motion v = u + at Here , v = 0 m/s u = 10 m/s a = -10 m /s square t = ? Put the values in the equation 0 = 10 m/s + (-10 m/s square * t ) 10 m/s square * t = 10 m/s t = 10 m/s / 10 m/s square t = 1s As the time taken by the ball to cover 5m distance is 1 sec so it will take 1+1 sec = 2 sec to reach the thrower

I have a physic test tomorrow about this topic😭😭😭

Number 2 Using V2=U2+2as Convert 72km/h to m/s =72x1000/60x60= 20m/s V=0 U=20m/s S=10 a=? 0^2=(20)^2+2ax10 0=400+20a 20a=-400 a=-400/20 a=-20m/s^2 I hope Im able to help those that were so confused at first like me 🙏

Your explanation is more than enough! Your videos should be tagged by RU-vid for students 👏👏

The ball when at the top, it is in motion exactly as much as it is in motion, when it is at the bottom. That is because the ball is always in motion within space-time. All you can do is change its direction of travel of the ball within the 4D environment known as space-time.

Aap jaise logo ki bahut jaruart hai sir hum students ko. Yha to sab bas ratta marne me mahir bnate hai jo mujhse nahi hota. Pahli baar lga ki mai sahi jagah par time investment kr rha hu. U r The great Teacher.......

Legends are watching this 1 day before exam

1(10) 2(-20) 3a(5) b(1)

Q1) u= 0 a=2 t=5 v=? (1)v=u+at= 0+(2 x 5)= 10m/s Q2) a=?, u=20 m/s(72 km/hr- m/s), v=0, s=10 (3) s= v^2- u^2/ 2a 10= 0^2 - 20^2/ 2x (a) a=-20^2/ 2x 10 a= -400/200 a= -4m/s Q3) Use equation v=u +at to find time= 1/20 then use equation s=ut+1/2at^2= 1/4 or 0.25 m

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The problem I'm facing is I can't understand where to put these equations for solving problems can u explain it sir pls?

In question no. 2 and 3 time is not given so, we can Put Time= Distance/Speed instead of time. And both distance and speed is given in the questions.

Hello everyone. I am currently doing 9th term 2 examinations. even though this chapter isn't there in my term 2 examination, i had to study it as these formulas were to be used once again. so plz pay attention to this motion chapter.

Sir you are doing the best you can. Where have you been in all my school life? Physics is now a piece of cake for me. Thank you sir! Thank you Manocha Academy.

Thank oh sir for such a good explanation 1)10m/s 2)-2m/s^2 3)h=5m T=2s

Here, i am from grade 10 , among the students of class 9 😂😂

Q.1) Given : initial velocity = 0 m/s, acceleration = 2 m/s^2, time = 5 s. To Find : speed of the car. To find the speed of the car we can apply the 1st equation of motion i.e v = u + at, v = 0 + 2*5, v = 0 + 10, v = 10 m/s. Therefore the speed of the car after 5 s is 10 m/s. Q.2) Given : Initial velocity = 72 km/hr. Now we have to convert 72 km/hr in m/s i.e 72 * 5/18 = 20 m/s. displacement = 10 m. final velocity = 0 m/s. To find : retardation of the brakes. Now to find the retardation of the brakes we can apply the 3rd equation of motion i.e v^2 - u^2 = 2as, (0)^2 - (20)^2 = 2a(10), 0 - 400 = 20a, - 400 = 20a, - 400/20 = a, - 20 m/s^2 = a. Therefore the retardation of the brakes is - 20 m/s^2. Q.3) Given : Initial velocity = 10 m/s, acceleration due to gravity = - 10 m/s^2, Final velocity = 0 m/s. To find : height & time taken. Now to find the height reached by the ball we can apply the third equation of motion i.e v^2 - u^2 = 2as, (0)^2 - (10)^2 = 2(-10)s, 0 -100 = -20s, -100/-20 = s, 5 m = s. Therefore the height reached by the ball is 5 m. Now to find the time taken to come back to the thrower we can apply 1st equation of motion i.e v = u + at, 0 = 10 + (-10)t, -10 = -10t, -10/-10 = t, 1 sec = t. Now the time taken for the ball to reach the height of 5 m was 1 sec. So, the time taken for the ball to reach to the person who throwed the ball will be 2 seconds. Sir I have solved the top three questions kindly tell me the answers are right or wrong.

Sir I easily solved the 1st and 3rd Question using the Equations . But I was unable to do the 2nd one , could you explain it ? Also I have a doubt in how to explain the equations of motion graphically ? Please explain sir , you are explaining very well , it is quite helpful for me . Regards , Abhilash Kar IX

Does it basically mean that whatever information is given to us after the = sign, we have to check that whether all the info. given in the question is the value of everything after the = sign in every equation?

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This lecture helps so many students like me every day..thnku so much for making this..

sir answers 1.10m/s 2.A=20metre/sec square 3.distance= 5m,time=1sec

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the video is so good like everything but please try to make a short video cuz there are student like me who are trying to cover whole syllabus 2 days before my exams.😅

1) V=10m/s 2)a=20m/s² 3)maximum height is 5metets Time taken back to thrower is 2 seconds

Sir please make one video to solve the equation of motion

My test is tomorrow and you've explained this way better than my teacher. A big Thank you all the way from Zambia. I will let you know if I pass

My understanding just went from not having a clue what any of this meant, to being able to get it right most of the time. Great video. You are doing a great job providing all the lectures for free. I really appreciate that. Your content and video quality are really good. Thank you, you just gained a subscriber.

This is the best video for explaining for people who have exams in 12 hours

1. Ans=10m/s 2. Retardation of the bus= minus acceleration= -20m/s² 3. Maximum height= -5m Time taken to reach maximum height=1s and to come back to the thrower =1s. Total time for the motion =2s

Ngl this is very helpful as I am starting my grade 11 this year,thank you very much

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Q1 answer is 10m/s Q2 answer is-20m/s² Q3 answers are 1sec and 5m

Sir, Could you please upload the answers in *your comment description* cuz It would be extra useful, since its hard to look for the answers when ever in doubt ;)

1. final velocity (v)= 10m/s 2. acceleration (a) = -20m/s^2 3. i) time (t)= 1s ii) displacement (s)= 5m

Crystal clear explanation sir! An apt example of experiential learning. I am a chemistry teacher but physics has a special place in my life.

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Respected Sir Physics and math are the hardest subjects for me. This was the video I was looking for. THANK YOU SO MUCH SIR. I request you to make more videos on physics, not only for me but for all the viewers of your wonderful teaching!!!👍😀

Sir can u plz reveal the ans.. I need to check my ans..

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Q1) v=u+at u=0, a=2m/s ,t=5s v=o+2×5=10m/s

sir i am haveing problem in 2nd and 3rd question plzz can you help me

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Sir you are great tomorrow I have a science periodic test in which motion is coming I have not studied at all and when I see you video I shocked and now I am ready to give tomorrow test thanks a lot

You are doing a great job providing all the lectures for free. I really appreciate that. Your content and video quality are really good. Thank you, you just gained a subscriber.

Legends are watching 1 hour before exam in 2x speed 😂

HOW TO FIND THE EXACT EXACT EXACT CIRCUMFERENCE OF CIRCLE. Procedure Step 1. Draw a square, 2 diagonals and inscribe a circle in the square. Thereby side and diameter will be the same Step 2 : Subtract 2 diagonals from the perimeter of square Step3. Divide step2 with 8 Step4. Add 3 times of the side. to Step 3 At the end we get the EXACT circumference of the inscribed circle

Not just Better than byjus , vedantu , topper .. but also effective and hardworking ...

I'm studing in 9th standard I have test tomorrow this video helped me thank you

Dear Sir, Can you please explain your audience the meaning of the equations of motion. Like, why we took out the half of acceleration times displacement and etc. This would be very helpful if you will make a separate video on this topic. Regards, Myth

ALL CONCEPTS CLEAR 😀😀 YOU GOT MILLIONS OF VIEWS INSHAALLAH☺☺

Sir I have doubt in 3rd question please give explanation of 3rd question it's urgent

a.) 10 m per second b.) 5 m per second per second c.) time to get to top = 1 sec Time taken to come back to the thrower= 2 seconds. Really appreciate your efforts Sir. Thanks.

1.v=10m/s 2.a=-20m/s^2 3.h=5m t=2s

Lack of quality of education in offline made a graduate to reach here and u proved my destination is perfect one to learn physics concepts thanks alot

If upward motion v = 0 then Q. Calculate v after 5 metres how v = 2.24m/s

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Sir may I ask how we've arrived at the formula v²=u²+2as ? I get that a is the change in v per time, but how did we get from a=v-u/∆t to v²=u²+2as?

Ultra legends are watching this 1 hour before the exam

Q1: Ans: Speed of the car after 5s is 10 m/s Q2: Ans: -20 m/s2 Q3: Ans: Maximum height reached by the ball is 5 m Time taken to come back to the thrower is 2 sec

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Teacher my answer are; Qn 1=10m/s Qn2,retardation =20m/s^2 Qn3;max height=5m Total time=2s Please teacher confirm if my solutions are correct. Thanks for all the videos, really helpful.

Sir Can you please upload a small video for giving the concept of third equation of motion? Thank you.

This is one of the best teachers In the world he takes time on his explanations, examples and he puts the examples into real life experience which makes it easier for the student watching to learn and retain for a very long time🙌

Sir mujhe puchna hai ki jo 10th cbse all subject course ke liye usme pura a to z chapter complete karte hai kya...

The answer is 1=10m/s 2=20m/s 2 3=5m

Sir can you explain what is the difference between Newton's Law of Motion and Newton's equation Of motion.

Thanks very much Sir for your clear and understandable way of teaching, this has really contributed more in my knowledge in science

It's really good sir .. Who are study persons they watch this video. Really it's good guys watch this .

Thank you very much Sir. Nicely explained. Is there a video for graphical representation of linear motion? Also it would be good if u could also add videos on numericals?

Sir you are really the best teacher

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Answers: 1) 10 m/s 2) -2 m/s2 ( due to retardation) 3) maximum height: 5m Time taken to come back to the thrower: 1s

Sir your method is awesome...that's how I wanted to learn

Sir, i think it's best if u put the answers in the comment immediately after the lesson so that we can check if our answers are correct.

Q1: 10 m/s Q2: I'm stumped on this one Q3: 10 m, 2 seconds

1. 10m/s 2.20m/s 3.hmax = 5m and and time it reaches back to thrower is 2sec

Please could you make a video solving these equations.

1= 10m/s 2= 20m/s square 3= 5m & 1s

Yaaa great to see the motion from the ball, initial velocity, from the top and the bottom . I see carefully from the speed of the motion of the ball of the velocity of the ball around of it.

Thank u so much sir for this amazing session....things are crystal clear Answers to the H.W questions 15:35 are :- 1.)10m/s 2.)-20m/s^2 3.)s=5m and t=1s

Q1: v=10m/s Q2: a=-20m/s^2 Q3: s =5m and total time =1s+1s =2s

Please explain question 2 and 3

1. 10m/s 2. Doubt 3. i) 5m ii) Doubt

better that 1 hour in school

Sir, is the free fall of the weightless leaves are same 9.8 or 10 m/s gravity?

You teach with good technic.

Sir I felt you are best in explaining topics about physics, I clearly understand your teaching sir, thank you sir🙏

I have a physics exam in like a week while I'll be writing other subjects so I wanna get these equations in my head