something very interesting in this video is even though u know how to solve the questions but you ask them can we or why don't we consider like you have no idea then you show them that they can do it. great teacher.
@@BicenMaths yes I say that now however when I become very successful person you are 1 of the people who helped me and I will talk about you in public. God bless
For the last question why would you want their initial velocity to be the same. You went from AC and AB. Would there be a difference if I go to AC and BC. If so what's the logic behind it, thank you
As the 2 journeys I am examining are starting at A, they must logically have the same starting velocity. If you did AC and BC they wouldn’t share the same starting velocity, as the journeys are being considered from different start points.
Hi Sir, first of all, thanks for your videos, they are very helpful. Secondly, I have a question regarding the last question on the video. I try to attempt the questions before your solution but I did it using the average velocity = displacement/time equation. I figured out that the average speed from A to B is 11m/s and the average speed from B to C is 26 m/s. Then using the definition of average speed, I figured out that the average speed/velocity corresponds to the speed at the midpoint of time. Using this principle, I figured out that the car travels at 11m/s at the midpoint of AB, hence at the time interval of 1 second and the car travels at 26m/s at the midpoint of BC which would be 4 seconds(2 + the midpoint of 4). So the car travels at 11m/s at t = 1 and at 26m/s at t = 4, so using the definition of acceleration, I did the change in velocity over the time interval so 26-15/4-1 which is 5m/s2 which is the same answer as using the simultaneous equations and furthermore the starting time must then be 11-5 (transition from t =1 to t=0 using the acceleration) to get 6m/s as the starting speed. I would like to know if this was a coincidence or this could be used as a legitimate method to solving these equations. (The average velocity thing can be explained using a velocity-time graph, where a diagonal line represents the acceleration. The area/ time would give us half of the velocity of the line, which would occur at half the time of the length of line. )
I believe this approach works, but it won't be something that I would recommend as you progress through - especially as in future chapters there will be variable acceleration, where some of your ideas won't hold up (for example, that the average speed is at the midpoint). Logically I think this approach does work, but try and stick to more conventional methods - only because they tend to repeat themselves quite a lot, so will save you time! :)
The direction of the object is shown by whether the velocity is positive or negative - if acceleration is negative it just means that it is slowing down, or getting faster in the negative direction!
Oh because the 20 was representing the displacement, i.e. it was 13t - 2t^2 = s. 20 was the specific displacement at a given time, but I was interested in the graph of the displacement, i.e. the graph of s
I'm in 2 minds about this... one, it has come up before, so perhaps less likely as it's something you can repeat. Two, it has come up before, so the examiners might enjoy asking about it! I'd probably gamble on it not being very likely, personally, but that's just my opinion!
well explained.100 times better than my physics teacher.
Great video Mr Bicen thank you so much!
Absolutely Godsent you are mate
something very interesting in this video is even though u know how to solve the questions but you ask them can we or why don't we consider like you have no idea then you show them that they can do it. great teacher.
Thank you! That’s a nice detail to have picked out - very perceptive of you.
@@BicenMaths yes I say that now however when I become very successful person you are 1 of the people who helped me and I will talk about you in public. God bless
You’re too kind! 🙌🏼
just binging yet another playlist even though i am aqa
same, but the way he teaches and how interactive these lesson are is just too great
For the last question why would you want their initial velocity to be the same. You went from AC and AB. Would there be a difference if I go to AC and BC. If so what's the logic behind it, thank you
As the 2 journeys I am examining are starting at A, they must logically have the same starting velocity. If you did AC and BC they wouldn’t share the same starting velocity, as the journeys are being considered from different start points.
Hi Sir, first of all, thanks for your videos, they are very helpful. Secondly, I have a question regarding the last question on the video. I try to attempt the questions before your solution but I did it using the average velocity = displacement/time equation. I figured out that the average speed from A to B is 11m/s and the average speed from B to C is 26 m/s. Then using the definition of average speed, I figured out that the average speed/velocity corresponds to the speed at the midpoint of time. Using this principle, I figured out that the car travels at 11m/s at the midpoint of AB, hence at the time interval of 1 second and the car travels at 26m/s at the midpoint of BC which would be 4 seconds(2 + the midpoint of 4). So the car travels at 11m/s at t = 1 and at 26m/s at t = 4, so using the definition of acceleration, I did the change in velocity over the time interval so 26-15/4-1 which is 5m/s2 which is the same answer as using the simultaneous equations and furthermore the starting time must then be 11-5 (transition from t =1 to t=0 using the acceleration) to get 6m/s as the starting speed. I would like to know if this was a coincidence or this could be used as a legitimate method to solving these equations. (The average velocity thing can be explained using a velocity-time graph, where a diagonal line represents the acceleration. The area/ time would give us half of the velocity of the line, which would occur at half the time of the length of line. )
I believe this approach works, but it won't be something that I would recommend as you progress through - especially as in future chapters there will be variable acceleration, where some of your ideas won't hold up (for example, that the average speed is at the midpoint). Logically I think this approach does work, but try and stick to more conventional methods - only because they tend to repeat themselves quite a lot, so will save you time! :)
hi, why is the acceleration not positive ( why has the sign not changed) because the object is moving in the opposite direction.
The direction of the object is shown by whether the velocity is positive or negative - if acceleration is negative it just means that it is slowing down, or getting faster in the negative direction!
For the last question, why don’t I go from A to B then B to C?
This could work too, just would have different calculations.
For one of the questions, why do did you use the 13t -2t^2 when showing the properties of the graph, and not -2t^t + 13t - 20 ! thank you
Oh because the 20 was representing the displacement, i.e. it was 13t - 2t^2 = s. 20 was the specific displacement at a given time, but I was interested in the graph of the displacement, i.e. the graph of s
Is deriving SUVAT formulae likely to come up in the exam? (Edexcel spec)
I'm in 2 minds about this... one, it has come up before, so perhaps less likely as it's something you can repeat. Two, it has come up before, so the examiners might enjoy asking about it! I'd probably gamble on it not being very likely, personally, but that's just my opinion!