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...
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
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.
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
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
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 🙏
Won't the time be 2 seconds? The time it uses to reach maximum height is the same time it will use to travel back from maximum height to the thrower, right?
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.
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
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?
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.
Thanks a lot :) 1st and 3rd answers (a and c) are correct!! Try the 2nd question again, retardation should be 20 m/s2. 3rd Question asks to calculate maximum height also. Let me know if you have any doubts!
All the best for your Physics test!! You can also watch this related video on Motion Sums and Graphs if it is helpful for your test: ua-cam.com/video/VEfccaYUFZY/v-deo.html
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
explanation of the 2nd question: Given: final velocity (v) = 0 (since the vehicle comes to rest) initial velocity (u) = 72 km/h => 72 x 5/18 = 20 m/s displacement = 10 m to find retardation (negative acceleration), v² = u² + 2as =>(0)² = (20)² + 2(a x 10) => 0 = 400 + 2 x 10a => 0 = 400 + 20a => 20a = -400 => a = -400/20 => a = -20 therefore the acceleration is -20 m/s
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Solution for Question 1: Given, Initial Velocity (u)= 0 Acceleration (a) = 2m/s² Time (t) = 5 seconds Final Velocity (v) = ? We know, v = u + at v = 0 + (2m/s²×5s) v = 10m/s Thus, the speed of the car after 5 seconds is 10m/s.
@@j.os_h Here, we need to find Final Velocity. Note that Final Velocity has the same sign as speed (m/s). Thus, we write it as 10m/s. We write the magnitude of Accleration as m/s². Therefore, if you write any value as m/s², then the value you are referring is acceleration. Example, 10m/s².
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.
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
Sir You are literally very inspiring and your teaching is out of this universe. I have subscribed to you since 5 years and i have been a very big fan of you. I wish you would have millions of views and subscribers Thank you sir
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
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
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.
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.
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.
@@Prideofgeminis Time is always positive so whenever you find that your answer has time in negative, know that you've done sth wrong in the steps. Maybe u used gravity acceleration in positive not negative.
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!!!👍😀
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.😅
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.
Answers : 1 = 10m/s 2 = - 2 m/s^2 3 = Height : 5m Time : 1 sec Sir please check my answers. Thanks for your efforts of making videos , makes our studies simple and intresting !
@@lijasarma33 u=10m/s V=0m/s S=? T=? a=-10m/s^2(because to find the time when the ball comes downward) To find T V =u + at 0=10+(-10× t) 0=10-10t 10t=10 t=1s To find s(distance) s=v^2-u^2÷2a S=0-100÷2×-10 s=-100÷-20 s=5m
(3)u=10 v=0 (because the ball comes to a momentary stop on reaching the max. height) g=10ms^-2 v^2=u^2-2aS 0=100-2×10×S 20S=100 S(max height reached)=5metre. v=u-at 0=10-10t 10t=10 t(time taken to reach max height)=1sec. the ball takes the same time to cover both upward and downward displacement, provided acc due to g is constant. Total time taken to reach thrower hand=1+1=2sec.
Nice bro.i undertand the first equation well.i will correct the mistake i ave done.for the second part can u elaborate more pls cus it says THE TIME TAKEN TO COME BACK TO THE THE THROWER.tnx once more
Here is the solution of Q3:As I discussed in the video you need to split motion into 2 parts: upward motion & downward motion since upward motion is retardation and downwards is acceleration. If you start with upward motion, u = 10 m/s (given), a = -10 m/s2. What is the velocity at top? Zero. So v=0 m/s. You need height. So that's 's'. So best equation is v2 = u2 + 2as. On solving you get, s = 5 m. This is max height. To find time, we need total time. So it will be time of upward motion + time of downward motion. For upward motion, you know u, a and v, so you can use v=u +at. t = 1s. Since time of upward motion is same as downward motion. So final answer is 2s.
Q1: u=0 m/s a=2 m/s2 t=5 second V=u+at =(0+2×5) m/s =10 m/s (Answer) Q2: Let u=72 km/hr =(72×5/18) m/s =20 m/s. v=0 m/s (Since bus comes to rest after applying brakes) v2=u2+2as 0×0=20×20+2×a×10 0=400+20a 400+20a=0 20a=-400 a=-(400/20) a=-20 m/s2. Since retardation is negative acceleration, Retardation=20 m/s2 (Answer) Q3: u=10 m/s v=0 m/s a=10 m/s2 v2=u2+2as 0=100+2×(-10s) 0=100-20s 100-20s=0 -20s=-100 s=5 metre (Maximum height reached) v=u+at 0=10+(-10t) 0=10-10t 10-10t=0 -10t=-10 Time of ascent is 1 second Since time of ascent is equal to time of descent so Total time taken to come back to thrower is (1+1)second=2 seconds
Q2 Here is the solution: Initial velocity u = 72 km/hr. But we need to convert it to m/s(SI units) since other data in sum are in SI units. So u = 72 x 5/18 = 20 m/s. Since bus comes to stop v = 0. s = 10 m. a = ? To find 'a' we need to use the equation v2 = u2 + 2as Substituting, we get a = -20 m/s2. Negative sign since it is retardation. So retardation = 20 m/s2. (we can remove the sign in final answer since are using the word retardation not acceleration). Q3) As I discussed in the video you need to split motion into 2 parts: upward motion & downward motion since upward motion is retardation and downwards is acceleration. If you start with upward motion, u = 10 m/s (given), a = -10 m/s2. What is the velocity at top? Zero. So v=0 m/s. You need height. So that's 's'. So best equation is v2 = u2 + 2as. On solving you get, s = 5 m. This is max height. To find time, we need total time. So it will be time of upward motion + time of downward motion. For upward motion, you know u, a and v, so you can use v=u +at. t = 1s. Since time of upward motion is same as downward motion. So final answer is 2s. I have discussed some graphical stuff in this video: ua-cam.com/video/VEfccaYUFZY/v-deo.html
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
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
____________ Example #1: A car starts from rest and has a uniform acceleration of 2 m/s^2. Find the speed of the car after 5 s. Solution #1: u - 0; a - 2 m/s^2; t - 5 s; v - ? v = 0 + 2 * 5 = 10 ____________ Example #2: A bus is travelling at 72 km/hr. Brakes are applied and the bus comes to a stop after 10 m. Find the retardation of the brakes. Solution #2: a - -10 m/s^2; s - 10 m; u - 72 km/hr - 20 m/s; v^2 - ? v^2 = 20^2 + 2 * (-10) * 10 = sqrt160 = 12.64 m/s^2 ____________ Example #3: A ball is thrown upwards with a velocity of 10 m/s. Find the maximum height reached by the ball and the time taken to come back to the thrower. (g = 10 m/s^2). Solution #3: Maximum height reached = H = u2 / (2 g) H = 10^2 / 2 * 10 = 5 m Time for upward movement and downward movement = u /g t = 10 / 10 = 1 s + 1s = 2 s (total time)
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Thank s
Sir can I contact you
thank you sir very well explained in easy manner
My son did this: V= 10m/s
Retardation= 259.2Km/hr
The third one, t=1s, S= 5m
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...
Can you send 2 and 3 question answer also
Did u got the ans
Can I get 2. I'm confused
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 too
Got same answer too
You're asked to find the speed!
Why did you use 20m/s when it isn't in the question above
Please explain
@@serifatayotunde8385 when you convert 72 km/h to m/s , the answer is 20 m/s.. m/s is the SI unit so we can't use km/h
We need teachers like this in school
Yes
@Ashwanth A S What do you mean ??
Absolutely....right
Kaisa laga Mera mazak 😂😂😂
Yeah
Anyone 2024?
Yes me .. class 9th
Yes me , class 9th
🔥
Yes, me at college 😄
Yessss
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.
Thank you so much I got confused
@@sk14_yt44 You are Welcome.
Hey! What made you use the 3rd equation? Especially since we were finding retardation. What does v2 mean? Final Velocity squared?
@@jaronrallos3844 I used 3rd equation of motion because due to that formula we can find the answer in fraction of sec.
Op bhai 💗
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.
Happy to hear that video was helpful :) Do check out more videos at www.manochaacademy.com
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
Aayein
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
2nd question 2nd part
-2m/s
is the answer....
thanks ,your answers were quite helpful..
@@pankajsood9089 -400/20=-20m/s² is the ans
but he has written -400/200,according to that the answer is -2..
\
Second question's answer is -20m/s2
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
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 UA-cam for students 👏👏
Legends are watching this 1 day before exam
Mee too😂😂
Haha I am
Me not
@@tanishqdokewar2233 oh
😁😁😁😁
1. final velocity (v)= 10m/s
2. acceleration (a) = -20m/s^2
3. i) time (t)= 1s
ii) displacement (s)= 5m
Won't the time be 2 seconds? The time it uses to reach maximum height is the same time it will use to travel back from maximum height to the thrower, right?
@@kimsimby9615 yes it will be 2s
1(10)
2(-20)
3a(5)
b(1)
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.
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
Excellent!! All your answers are correct :)
Can u explain the answer for third question 🤔
How s =5m????
@@shreevigneswaran2112
U=10m/s
S=?
a= -10m/s^2
t=?
Here we use
V^2=U^2 +2as
0=(10)^2+2×-10 ×s
0=100 -20s
20s=100
S=100/20
S=5m
I hope helpful 👍
1 answer is 10m/s
2 answer is-20m/s^2
3 answer is t=1s ,h=15m
Thank oh sir for such a good explanation
1)10m/s
2)-2m/s^2
3)h=5m
T=2s
Hlo sis how did you get these answers
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?
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.
Thanks a lot :) 1st and 3rd answers (a and c) are correct!! Try the 2nd question again, retardation should be 20 m/s2.
3rd Question asks to calculate maximum height also.
Let me know if you have any doubts!
I got the second one correct sir
I was searching the ans of 2 thank you sir '
I have a physic test tomorrow about this topic😭😭😭
All the best for your Physics test!! You can also watch this related video on Motion Sums and Graphs if it is helpful for your test: ua-cam.com/video/VEfccaYUFZY/v-deo.html
Me tooooo
I have physics and chemistry
Same 😞
same😕
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
i got -20 m/s^2
explanation of the 2nd question: Given:
final velocity (v) = 0 (since the vehicle comes to rest)
initial velocity (u) = 72 km/h => 72 x 5/18 = 20 m/s
displacement = 10 m
to find retardation (negative acceleration),
v² = u² + 2as
=>(0)² = (20)² + 2(a x 10)
=> 0 = 400 + 2 x 10a
=> 0 = 400 + 20a
=> 20a = -400
=> a = -400/20
=> a = -20
therefore the acceleration is -20 m/s
a=-20m/s2
yh but they are asking retardation which is negative acceleration, which is again 20 m/s
Sorry to ask I'm just confuse I know u =72 so how did that 72*5/18 got in please
Mind blowing Sir u teach In a very polite manner
Thanks and welcome
This lecture helps so many students like me every day..thnku so much for making this..
It's my pleasure
Sir please make one video to solve the equation of motion
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.
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.......
Solution for Question 1:
Given,
Initial Velocity (u)= 0
Acceleration (a) = 2m/s²
Time (t) = 5 seconds
Final Velocity (v) = ?
We know, v = u + at
v = 0 + (2m/s²×5s)
v = 10m/s
Thus, the speed of the car after 5 seconds is 10m/s.
Please explain why it is not 10m/s^2 ??
@@j.os_h Here, we need to find Final Velocity. Note that Final Velocity has the same sign as speed (m/s). Thus, we write it as 10m/s.
We write the magnitude of Accleration as m/s². Therefore, if you write any value as m/s², then the value you are referring is acceleration. Example, 10m/s².
1) V=10m/s
2)a=20m/s²
3)maximum height is 5metets
Time taken back to thrower is 2 seconds
Not just Better than byjus , vedantu , topper .. but also effective and hardworking ...
Glad to hear that.
15:35
Q1 ans = v = 10m/s
Q2 ans = -a = 20m/s²
Q3 ans = s= 5m
sir answers
1.10m/s
2.A=20metre/sec square
3.distance= 5m,time=1sec
please show me your workings
How did you get the answer of 3rd
Respected Sir u r a great teacher and I m ur fan. How I van attend ur Series/Chapter vise lectures on UA-cam. hope for swift response.
Thanks a lot :) Do check out more videos at www.manochaacademy.com
Please give me the date of live class on equation of motion
Super sir.......it is helping even for the students who are little slow.... Very well explained....🙏🙏🙏👍👍👍👍
7:56 , here 9.8 is being mentioned right?
Yes
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.
Expert teacher
Hats off to you sir 👌🏻🙏🏻
The problem I'm facing is I can't understand where to put these equations for solving problems can u explain it sir pls?
😞😞 exams monday
@@gameomania9551 all the best friend all wishes
@@ZenCho007 tq bro love u
@@gameomania9551 OK bro you study better OK
@@ZenCho007 sure
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
Sir
You are literally very inspiring and your teaching is out of this universe.
I have subscribed to you since 5 years and i have been a very big fan of you. I wish you would have millions of views and subscribers
Thank you sir
Sir i m extremly requesting you to make a video on thermal conductivity and if possible so plz cover the whole lesson of thermodynamics
Sir, is the free fall of the weightless leaves are same 9.8 or 10 m/s gravity?
I too having same doubt
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
COPY PASTE!
You are a born teacher. Your students must be happy and lucky to have you as their teacher just like us over here on UA-cam.
Glad you liked. Keep watching
Sir you are really the best teacher
Thank you sir, u helped me last year and I've made it going to University 😊😊😊😊
Ngl this is very helpful as I am starting my grade 11 this year,thank you very much
You gained a subscriber sir.
Really, no physics channel teaches as practically as you do. I understood well😊
Thanks! Great to hear that! Do share our channel with your friends!
True
Sir you are pro at teaching appreciate your teaching style
Thanks and welcome
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
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.
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.
Thanks for subscribing!! Do share it with your friends!
Q.1 ans- v =10m/s
Q.2 ans - a=-40m/s
Q.3 ans- s = 10m , t=1s
2nd ques ans is retardation (-ve acceleration) of 20m/s^2 or can say -20m/s^2
You are a m0r0n bro
Answer is -20m/s^2
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.
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
You've failed as you didn't inform😂??
Best teacher clear concept 👌🙌 and in less time
Thanks
Q1: 10 m/s
Q2: a=-20 m/s squared
Q3: s= 5m
t= 1 sec . Are the answers correct ?
Q3 is -5m and -1 sec
@@Prideofgeminis farida is correct
@@Prideofgeminis Time is always positive so whenever you find that your answer has time in negative, know that you've done sth wrong in the steps. Maybe u used gravity acceleration in positive not negative.
Q1) v=u+at
u=0, a=2m/s ,t=5s
v=o+2×5=10m/s
Correct!
Superb Sir... I am seeing this video in 2020 again... Salute to u
Thanks for watching again
@@ManochaAcademy Welcome sir
Please could you make a video solving these equations.
This channel need to be millions of views and subscriber
Wow ! Thanks
@@ManochaAcademy welcome....and continuously provide videos
Crystal clear explanation sir! An apt example of experiential learning.
I am a chemistry teacher but physics has a special place in my life.
This teacher like gold
Hai
Yes
Yes
Very good teacher if he is in my school I will get physics full marks
Not gold diamonds
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!!!👍😀
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.😅
Someone like me trying to cover up the whole syllabus 1 week before the exam 😂😂💔. Jamb no be my mate
Ans 1. v=10m/s
Ans 2.a= -0.03m/s2
Ans 3. s=5m ,t=1sec
Is this the answers of top 3 q ??
@@anandvardhansharma4761 yes
yess!! You are absolutely right
Okk.....thnx
1---- 10m/s
2------- -20m/s2
3____ 5m, t= 1s
Please tell me the formula for 2nd question.Is it s=ut+1/2at^2
It will be v^2 = u^2 + 2as
First#v=u+at, 2nd#v2-u2=2as and 3rd #H=u2/2g☺️👍
@@tokitahmidahmed13 right☺️
@@kusumbanaly3966 thanks
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.
this is what i got
Pls how did you solve the Q2?
Sir can you explain what is the difference between Newton's Law of Motion and Newton's equation Of motion.
Laws of motion are applicable for all kinds of motion, equations are only applicable for motion that has constant acceleration
Sir Can you please upload a small video for giving the concept of third equation of motion?
Thank you.
Yes sir
Yes sir
Answers :
1 = 10m/s
2 = - 2 m/s^2
3 = Height : 5m Time : 1 sec
Sir please check my answers.
Thanks for your efforts of making videos , makes our studies simple and intresting !
I think second answer is 20
Soory -20
@@muhammadfaizantariq7537 ur right he's wrong
Please explain the 3rd solution
@@lijasarma33 u=10m/s
V=0m/s
S=?
T=?
a=-10m/s^2(because to find the time when the ball comes downward)
To find T
V =u + at
0=10+(-10× t)
0=10-10t
10t=10
t=1s
To find s(distance)
s=v^2-u^2÷2a
S=0-100÷2×-10
s=-100÷-20
s=5m
If upward motion v = 0 then Q. Calculate v after 5 metres how v = 2.24m/s
Sir can u plz reveal the ans.. I need to check my ans..
If you throw a ball in space the ball will move unifrom motion until the unbalance force is apllied.❤
ALL CONCEPTS CLEAR 😀😀 YOU GOT MILLIONS OF VIEWS INSHAALLAH☺☺
Q1: v=10m/s
Q2: a=-20m/s^2
Q3: s =5m and total time =1s+1s =2s
Very good you will be successful
sir i am haveing problem in 2nd and 3rd question plzz can you help me
(3)u=10
v=0 (because the ball comes to a momentary stop on reaching the max. height)
g=10ms^-2
v^2=u^2-2aS
0=100-2×10×S
20S=100
S(max height reached)=5metre.
v=u-at
0=10-10t
10t=10
t(time taken to reach max height)=1sec.
the ball takes the same time to cover both upward and downward displacement, provided acc due to g is constant.
Total time taken to reach thrower hand=1+1=2sec.
isn,t its the formula V^2=U^2+2aS?
Nice bro.i undertand the first equation well.i will correct the mistake i ave done.for the second part can u elaborate more pls cus it says THE TIME TAKEN TO COME BACK TO THE THE THROWER.tnx once more
Sir I have doubt in 3rd question please give explanation of 3rd question it's urgent
Here is the solution of Q3:As I discussed in the video you need to split motion into 2 parts: upward motion & downward motion since upward motion is retardation and downwards is acceleration.
If you start with upward motion, u = 10 m/s (given), a = -10 m/s2. What is the velocity at top? Zero. So v=0 m/s. You need height. So that's 's'. So best equation is v2 = u2 + 2as. On solving you get, s = 5 m. This is max height.
To find time, we need total time. So it will be time of upward motion + time of downward motion. For upward motion, you know u, a and v, so you can use v=u +at. t = 1s. Since time of upward motion is same as downward motion. So final answer is 2s.
Q1: u=0 m/s
a=2 m/s2
t=5 second
V=u+at
=(0+2×5) m/s
=10 m/s (Answer)
Q2: Let u=72 km/hr
=(72×5/18) m/s
=20 m/s.
v=0 m/s (Since bus comes to rest after applying brakes)
v2=u2+2as
0×0=20×20+2×a×10
0=400+20a
400+20a=0
20a=-400
a=-(400/20)
a=-20 m/s2.
Since retardation is negative acceleration,
Retardation=20 m/s2 (Answer)
Q3:
u=10 m/s
v=0 m/s
a=10 m/s2
v2=u2+2as
0=100+2×(-10s)
0=100-20s
100-20s=0
-20s=-100
s=5 metre (Maximum height reached)
v=u+at
0=10+(-10t)
0=10-10t
10-10t=0
-10t=-10
Time of ascent is 1 second
Since time of ascent is equal to time of descent so
Total time taken to come back to thrower is (1+1)second=2 seconds
Wow!
@@tinotendachibanda9390 Thanks sir
Third Answer is wrong
@@atharvamehta987 plz u tell the correct
Best teacher ever!!
Very good explanation sir
please upload a video on graphical representation of equations of motion and please tell q 2 and 3
Q2 Here is the solution:
Initial velocity u = 72 km/hr. But we need to convert it to m/s(SI units) since other data in sum are in SI units.
So u = 72 x 5/18 = 20 m/s.
Since bus comes to stop v = 0. s = 10 m. a = ?
To find 'a' we need to use the equation v2 = u2 + 2as
Substituting, we get a = -20 m/s2. Negative sign since it is retardation.
So retardation = 20 m/s2. (we can remove the sign in final answer since are using the word retardation not acceleration).
Q3) As I discussed in the video you need to split motion into 2 parts: upward motion & downward motion since upward motion is retardation and downwards is acceleration.
If you start with upward motion, u = 10 m/s (given), a = -10 m/s2. What is the velocity at top? Zero. So v=0 m/s. You need height. So that's 's'. So best equation is v2 = u2 + 2as. On solving you get, s = 5 m. This is max height.
To find time, we need total time. So it will be time of upward motion + time of downward motion. For upward motion, you know u, a and v, so you can use v=u +at. t = 1s. Since time of upward motion is same as downward motion. So final answer is 2s.
I have discussed some graphical stuff in this video: ua-cam.com/video/VEfccaYUFZY/v-deo.html
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
Can u please explain tje 2nd question didnt understand that
This is the best video for explaining for people who have exams in 12 hours
Sir answer are
1) 10 m/s^2
2)259.2 m/s^2
3)1second
In Q2 how can we use Newtown eqn of motion cuz the acceleration is changing right
Yes we will use 3rd equation that is 2as=v^2-u^2 @@MrSureshchinnappa
Answers...
1. 10 m/s
2. 20 m/s^2
3. h=5m and time taken to return = 2 s
Eveey thing os correct
Q2 answer is -2
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
Happy to hear that you liked the video :) All the best for your test!!
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
Bhot hard
Sir your method is awesome...that's how I wanted to learn
Glad to hear that
You teach with good technic.
No extra bakwas only knowledge ......
I like this chanel....
Your English speaking is very clear that can be understandable
1.v=10m/s
2.a=-20m/s^2
3.h=5m t=2s
Sir! I have answered a question that is (que 1)
V=u+at^2
V=0+2m/s^2(into)5s
V=0+10m/s
So V=10m/s
Correct!
the equation has no power of 2
1. 10m/s
2.20m/s
3.hmax = 5m and and time it reaches back to thrower is 2sec
Hey
____________
Example #1:
A car starts from rest and has a uniform acceleration of 2 m/s^2. Find the speed of the car after 5 s.
Solution #1:
u - 0; a - 2 m/s^2; t - 5 s; v - ?
v = 0 + 2 * 5 = 10
____________
Example #2:
A bus is travelling at 72 km/hr. Brakes are applied and the bus comes to a stop after 10 m. Find the retardation of the brakes.
Solution #2:
a - -10 m/s^2; s - 10 m; u - 72 km/hr - 20 m/s; v^2 - ?
v^2 = 20^2 + 2 * (-10) * 10 = sqrt160 = 12.64 m/s^2
____________
Example #3:
A ball is thrown upwards with a velocity of 10 m/s. Find the maximum height reached by the ball and the time taken to come back to the thrower. (g = 10 m/s^2).
Solution #3:
Maximum height reached = H = u2 / (2 g)
H = 10^2 / 2 * 10 = 5 m
Time for upward movement and downward movement = u /g
t = 10 / 10 = 1 s + 1s = 2 s (total time)
Thanks you're an amazing person
In question 1 aren't supposed to be look for speed and not velocity?
@@emmanuellaayeni7301 Speed and velocity is the same except for their scalar and vector quantity
Q2, how did u know the acceleration
Sir from where are you I scored 70 out of 70 in my exams because of you only
Sir please make video of graphical representation of motion