Sir I'd just migrated to the UK from India straight into y12 and to make it worse id missed 8 months because of visa delay , the only reason I'm still performing well is because of your videos . Thanks a lot sir .(Pure and Further maths )
Great to hear that! Make sure you check out my Bicen Maths Google Drive (link in the About section) as it’s full of things you might find useful - like the blank pdfs of the lessons I teach, exam questions, etc. 👍🏼
Hi, thanks for this video! Just wanted to know will the question always tell us which type of graph to draw? And if the question involves calculating acceleration and deceleration, is it always a velocity-time graph?
They'll always say which type of graph, correct. Yes, that is also correct about it being velocity-time for acceleration! Velocity-time is the most common type by far.
at 37.41 in question 6 you used the formula of a triangle as bxh/2. but thats only for right angled triangles. that triangle with area of 1500 is not a right angled triangle. so how can u use that formula???
That formula doesn't only apply to right angled triangles - see here! www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html It applies when the base and height measured are perpendicular.
Hey! The textbook questions may refer to finding the displacement from starting point to __, but the object returns back ‘home’, so I put the displacement to be 0 but rather it was looking for the area under the graph. When would a question explicitly want that answer of 0 (when an object leaves and returns home)?
It might not have worked for you because the acceleration wasn't constant, and you were using SUVAT, which can only be used with constant acceleration - so if there's changes in acceleration, drawing a speed time graph is better, and then interpreting the distance travelled as the area under the graph!
hello sir, why did you convert the answers to 'b' and 'c' to km h instead of keeping it as km min at 13:42? i understand why you did in part a but that's because it was included in the Q.
I think we're assuming that speed/velocity should be given in a standard unit, which should be km/h. I imagine in an exam they would be clearer about this!
The actual maths will be very relevant, yes - luckily the “maths” is pretty much always the same. These topics are assessed in the same way across the exam boards, there are slight differences, mostly in statistical sampling and other small areas. I also use lots of Edexcel exam questions, but these will be great preparation for other boards too 👍🏼
I thought the Area under the graph showed distance - so why do we say its displacement - because in the second example wouldn't the particle end up at the same place so the displacement is 0?
It depends on our definitions of areas - if you allow area to be negative, then it would represent displacement, as you might get some curves which are above and below the axis, so the resulting area is a combo of the positive and negative bits, producing the displacement from where it started. If you take the bit below as positive, then you add the areas together, you would get the distance travelled.
Hey sir. Could u pls explain why area under a velocity time graph is distance and not displacement? Velocity is a vector, so the area would give a vector as well in terms of displacement, no?
Take an example where something travels in one direction, and then returns in the opposite direction at the same speed. Its displacement would be zero, right? But if you drew the velocity-time graph, it would be a triangle shape (with the peak where it reverses and comes back). Clearly the area under the triangle isn't zero (the displacement), and instead it represents the distance travelled. Hope this example shows why!
@@BicenMaths how would the graph be a triangle, wouldnt it be a graph with a rectangle above the graph and a rectangle below the graph with the same area as when the object goes in the opposite direction, velocity should be negative. This graph would also show a displacemnt of 0.
If you get a constant (i.e. just a number) then you have a straight line (imagine differentiating 5x + 3, you'd just get 5). If you got a function still, then it's variable.
aren't we supposed to divide by 60 in order to convert minutes to hours? I don't understand why the answer is multiplied by 60 to convert kmmin to kmh in the first question part c.
Think of km per minute meaning how many km you travel in one minute. Km per hour is how many km would you travel in an hour - which is 60 times longer, hence multiplying by 60.
@@BicenMaths oh I get it. I can only divide like that in the early stages of the question. Thank you so much, sir. For the videos and the explanation :D
