For 6:20 Him explaining the why triangles need 1/2 in front is because if you mulitply base and height you would get the points (4,0) (4,2) (6,0) (6,2) but the line cuts half of the rectangle is like a rectangle has two triangles you know so when you cut them half you'll get the triangle that's why the 1/2 is in the formula of finding are of a triangle ... Whatever he said was right but I'm just giving another method that is true too...
Initial velocity must be provided. It cannot be derived from an acceleration vs. time graph. He should have made that clear when he provided the information.
@@abbhijitsingh841 The velocity was assumed you cannot find it out through the acceleration graph normally i think it is assumed that velocity is 0 or its given
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
also to help, if you picture it, the triangle is just half of a square and the square is just half of a rectangle. so if you take the acceleration of the rectangle (8m/s^2) and divide that by 2, you get the acceleration of the square. and again, since the triangle is just half of the square, you can divide the acceleration of the square (4m/s^2) and divide that by 2 again. giving you 2m/s^2. so basically divide the rectangle acceleration by 4 since 4 triangles make up a rectangle.
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
Sir can u please explain why the velocity at 6 is greater than velocity at 4 sec?..from the graph we can see the object's acceleration is decreasing r8?..it means the final velocity must be lessened by time..but why in here the velocity at 6 is greater than at 4 sec?..it doesn't approve the graph then in my opinion...plz point out my mistake where I'm misunderstanding..
I had the same misconception at first, but you must take into account that, at time 6 the acceleration indeed has decreased, but it is not zero nor negative, so the velocity is still increasing, at a lower pace than before but still increasing.
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
An electric train moves from rest with a uniform acceleration of 1.5m/s>2 for the first 10s and continues accelerating at 0.5m/s>2 for a further 20s.it continues at constant speed for 90s and finally takes 30s to decelerate uniformly to rest.Draw a graph of speed against time for the journey. sir, how do i draw that graph?
try to draw the acceleration first, its a big graph so you might want to scale it. the acceleration is essentialy your slope, then you can perhaps do perhaps divide the graph in different areas, each for every different acceleration valu, and maybe try to integrate the different curves you find. that should give you a graph representing velocity, thats my guess tho i dont assure you anything
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
Sir, I would be very grateful if you answer this question: How is that a object accelerating at a speed of 9.8m/s but just covering 4.9m?Hope to get a big explanation. I am having a lot of trouble with this questions.So please kindly reply,Sir.
Hey, distance is basically speed times the time taken, in other words distance d=v*t if you plug the values 4.9 and 9.8 for d and v consecutively, you get time = 0.5 seconds. So your object basically just traveled at 9.8 m/s for 0.5 second and therefore covered just 4.9 meters of distance. Hope that helped.
It really helped me so much.Thank you very much.But it would be grate if you answer this question: A object falling due to gravity for 1second.And it,s acceleration is 9.8m/s^2. So why does it travel 4.9m instead of 9.8m??
@@jannatulnyeemratul6039 Hey, so the equation used for determining the traveled distance of falling bodies is h= ut + 1/2gt^2 where g is the gravitational constant which is basically 9.8 m/s^2. So assuming that the object we are considering here fell from a still position we'll plug 0 as our initial velocity that is u in the equation . Plug all the values into the equation and you'll get h = 4.9m. Hope that helped. Also, if you wanna know how this equation came into being I can give you links to videos that explain the whole thing.
Hmm, the first 4 seconds on the graph indicated that Daisy is moving at a constant velocity of 2m/s, how ever what you got wrong is that there is no initial change in acceleration in the first 4 seconds...
That is the exact mistakes I made about 2 hrs ago. It took me like 15 minutes to find my mistake. The thing is, the vertical axis representing the acceleration, not the velocity.
I'm sorry, but this video just confused everything I just understood from Sal Khan's videos.. Please just get to the point, stop with the "but wait a minute stuff". Just tell us whats going on and why. No need to dance around it.
youre confused its because you dont know the basic concepts of math, try to start from the very basics of math instead of jumping topic from a to z like from arithmetic to advanced calculus quickly
Provided an initial velocity which was v = 1 m/s find the final velocity at t = 4, recall acceleration = Δv/t , hence Δv =at, Δv = final velocity ﹣ initial velocity. You know Δv = 8 m/s at t = 4 and you've been given initial velocity, so you can find out the final velocity = Δv + initial velocity = 1 + 8 = 9 m/s
love your work, reallly helping me out thanks again
Sir..🙏🏻🙏🏻🙏🏻😌 thank u very much
I am in love with Khan Academy
You know this is a hard graph when Khan Academy comes first in the search list
The best teaching ❤
How do you calculate distance from a acceleration-time graph?
Thank you! This really helps alot for my exams lol
For 6:20 Him explaining the why triangles need 1/2 in front is because if you mulitply base and height you would get the points (4,0) (4,2) (6,0) (6,2) but the line cuts half of the rectangle is like a rectangle has two triangles you know so when you cut them half you'll get the triangle that's why the 1/2 is in the formula of finding are of a triangle ... Whatever he said was right but I'm just giving another method that is true too...
Thanks bro
Love your work and good one on the jerk😅😅😅😅
Thanks that was very interesting! Just make sure to use vector notation where needed
They was sooo helpful ❣️
did you just assume the gender of the instructor
thank you. never heard of jerk before
well explained
how about displacement
which software is used in making this video
@7:20, how did you get V=1 m/sec when T=0?
Initial velocity must be provided. It cannot be derived from an acceleration vs. time graph. He should have made that clear when he provided the information.
@Omran Alriahi he made it up since it is not shown in an acceleration vs. time graph
@Amurie_UA-cam how did you get it?
pls help...
@@abbhijitsingh841 The velocity was assumed you cannot find it out through the acceleration graph normally i think it is assumed that velocity is 0 or its given
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
also to help, if you picture it, the triangle is just half of a square and the square is just half of a rectangle. so if you take the acceleration of the rectangle (8m/s^2) and divide that by 2, you get the acceleration of the square. and again, since the triangle is just half of the square, you can divide the acceleration of the square (4m/s^2) and divide that by 2 again. giving you 2m/s^2. so basically divide the rectangle acceleration by 4 since 4 triangles make up a rectangle.
Does the initial velocity always mean 1? I thought its supposed to be zero
Vi has to be given as you cannot get it from the graph. He made it up.
Do I need a Vi or Vf all the time with an Acceleration vs Time graph?
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
Sir can u please explain why the velocity at 6 is greater than velocity at 4 sec?..from the graph we can see the object's acceleration is decreasing r8?..it means the final velocity must be lessened by time..but why in here the velocity at 6 is greater than at 4 sec?..it doesn't approve the graph then in my opinion...plz point out my mistake where I'm misunderstanding..
I had the same misconception at first, but you must take into account that, at time 6 the acceleration indeed has decreased, but it is not zero nor negative, so the velocity is still increasing, at a lower pace than before but still increasing.
David Herrero Thank u so much😊. Well I had figured out myself later😅. Well thanks again bro.
Why is the change in velocity at 0 = 1
Nvm, he assigned a value
Great video though
YA! that threw me off too.
@@liamprior9508 same
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
Can you do more videos of acceleration and the different graphs
Thanks
Sir can we make v/t graph from a/t then find v max
An electric train moves from rest with a uniform acceleration of 1.5m/s>2 for the first 10s and continues accelerating at 0.5m/s>2 for a further 20s.it continues at constant speed for 90s and finally takes 30s to decelerate uniformly to rest.Draw a graph of speed against time for the journey.
sir, how do i draw that graph?
try to draw the acceleration first, its a big graph so you might want to scale it. the acceleration is essentialy your slope, then you can perhaps do perhaps divide the graph in different areas, each for every different acceleration valu, and maybe try to integrate the different curves you find. that should give you a graph representing velocity, thats my guess tho i dont assure you anything
o.m.g! thank you
I dont get the part of: V at 0 = 1 7:10
He made it up, since it cannot be solved with it
@@federalbureauofinvestigation4 thx
well by formula a=(v-u)/t where v=final velocity and u=initial velocity. a=2m/s^2 from graph, v is the area under that is 8m/s and t=4sec. by this we get u=1m/s.
2? Not -2?
So the people who are referred to as jerks are changing their acceleration with time and that annoys other people
Sir, I would be very grateful if you answer this question: How is that a object accelerating at a speed of 9.8m/s but just covering 4.9m?Hope to get a big explanation.
I am having a lot of trouble with this questions.So please kindly reply,Sir.
Hey, distance is basically speed times the time taken, in other words distance d=v*t if you plug the values 4.9 and 9.8 for d and v consecutively, you get time = 0.5 seconds. So your object basically just traveled at 9.8 m/s for 0.5 second and therefore covered just 4.9 meters of distance. Hope that helped.
It really helped me so much.Thank you very much.But it would be grate if you answer this question:
A object falling due to gravity for 1second.And it,s acceleration is 9.8m/s^2.
So why does it travel 4.9m instead of 9.8m??
@@jannatulnyeemratul6039 Hey, so the equation used for determining the traveled distance of falling bodies is h= ut + 1/2gt^2 where g is the gravitational constant which is basically 9.8 m/s^2. So assuming that the object we are considering here fell from a still position we'll plug 0 as our initial velocity that is u in the equation . Plug all the values into the equation and you'll get h = 4.9m. Hope that helped. Also, if you wanna know how this equation came into being I can give you links to videos that explain the whole thing.
Thanks a lot for your help.It really helped me a lot.Again thanks,Iftekhar Ahamad.
@@iftekharahamad5690 legit chad. we need more people like you.
how do we find the Avg acceleration from the acceleration time graph
looking for this too. did you find out how?
Maybe total acceleration devided by total time?
What program is this?
that's a fast dog
Indeed
*J*
*O*
*L*
*T*
Hmm, the first 4 seconds on the graph indicated that Daisy is moving at a constant velocity of 2m/s, how ever what you got wrong is that there is no initial change in acceleration in the first 4 seconds...
That is the exact mistakes I made about 2 hrs ago. It took me like 15 minutes to find my mistake. The thing is, the vertical axis representing the acceleration, not the velocity.
I'm sorry, but this video just confused everything I just understood from Sal Khan's videos.. Please just get to the point, stop with the "but wait a minute stuff". Just tell us whats going on and why. No need to dance around it.
youre confused its because you dont know the basic concepts of math, try to start from the very basics of math instead of jumping topic from a to z like from arithmetic to advanced calculus quickly
he explained so well, he goes to the very root in order for you to make sense what is happening🙂
@@newfire2633 yeah true but too slow
are u indian
Viraj Murab he is Bengali
Surely V4 - 1(m/s), = 8(m/s) - 1(m/s) = +7(m/s)? not 9(m/s). Might be me being stupid.
Provided an initial velocity which was v = 1 m/s find the final velocity at t = 4, recall acceleration = Δv/t , hence Δv =at, Δv = final velocity ﹣ initial velocity. You know Δv = 8 m/s at t = 4 and you've been given initial velocity, so you can find out the final velocity = Δv + initial velocity = 1 + 8 = 9 m/s
I am very surprised by the Jerk
So the people who are referred to as jerks are changing their acceleration with time and that annoys other people