Sir I just really want to thank you. I found your videos yesterday and they're really thorough and you explain things so simply it enables me to understand stuff, and the little tips and tricks you give too are great. My stats and mechanics teacher is a member of Mensa so he explains stuff as if it was complex when really its just so simple, and cuz im in the "top set", he caters the lessons towards the geniuses that should've taken further maths but didn't want to. So genuinely, thank you so so so much !!! I have so much more faith in myself, and am feeling so much more confident, these are literally amazing !!!!
Hi sir, just wanted to thank you for all the help you have given me and others in your a level maths videos. You always teach at a pace where beginners can pick up what you're teaching and that's really helpful for me, as sometimes my teachers teach so fast that's it's difficult for me to sometimes process what I've learnt. You are a really amazing teacher! Thanks for all the help that you've given me and others, once again :) wish you all the best!
This is such a kind thing to say! I really appreciate you taking the time to write this to me - thank you. Wishing you all the best with your studies - let me know if there’s anything I can do to help! Good luck!
hi sir, i am currently in year 12 studying a level mathematics, and your videos have had a stupendous effect on my learning. hence, i give you my sincerest thanks and wish you all the best for your future aspirations in life. bye!
at 14:20, the overall length of the journey consists of -9.8 acceleration in the first half and 9.8 acceleration in the second half, how do you know which acceleration value to use?
Hi, at time stamp 14:33 why is acceleration still positive? You said that in this example, going down is positive. However, in the question, to get to the greatest height we are required to go upwards from the starting position, so shouldn’t the acceleration be -9.8? And so can’t the acceleration just be 21? Many thanks you’re incredibly helpful!
You set the direction for the whole question - and in this example, anything that is downwards is positive, hence acceleration being positive. As the stone is initially moving upwards with speed 21, it is -21, as upwards is therefore in the negative direction. It doesn't matter that the stone eventually goes downwards, at the start it is going up, hence u = -21
27:15, why is the point t2=3.498 refer to the time where it starts going straight down? not the total time, like how in the previous example. 4.6 referred to the total time (and 0.31 refers to the time it would take to get from the ground to the starting point)
The times are referring to when it is 10 metres above the starting point, so it makes sense that the first time is referring to its journey up, and the second time is its journey downwards!
Yep! That’s absolutely correct. You can pick either direction to be positive - as long as you indicate the ones in the opposite direction as negative, the formulae work!
If the ball is going up, but it is accelerating downwards, this means that the ball is not getting faster, but is actually getting slower. If you say that up direction is positive, then acceleration is downwards, and so is therefore negative - negative acceleration means that its speed is DECREASING, not increasing, which is in line with how a ball's speed changes as it moves upwards against gravity!
at timestamp 9.24, why isn't velocity -5.2 instead? Because I took upwards as my positive, doesn't it mean that velocity in opposite direction is negative ? Also, when square rooting to get rid of v^2 there is a possibility of 2 values. Why is it the positive and not the negative one?
You have to interpret it in the context of the question to decide whether it is moving in the positive or negative direction - you’re right that the square rooting leads to 2 answers, so you need to select which one carefully by thinking about the mechanics of the situation.
hello sir, this may be a silly Q but at 30:05 why would displacement be 0 if the blue dot is not at the same point as the black dot.. i thought displacement took direction into account so wouldn't the value of s be equal to that distance. OR is the value of s, 0, because both dots are at the same 'level'
s is 0 as they are both at the same level! You might hear me say this in some videos, but I usually think of displacement as taking a photo at the two points you're considering, and just saying how the two photos differ. In this case, the photos would look identical, so there's no displacement!
So you could do either, but by doing it from when it is projected, I can calculate the time in one single calculation. If you do it from the top back down again, you'd also need to think about from when it was projected to the top (which in this case is very easy, as it just the same time up and the same time down!).
Sir, hope you had a great Christmas and have a happy new year I don't understand the concept of the arrow determining whether to use + or - for speed, acceleration etc like the one you put in a circle. Whenever you see this comment could you please explain this idea again as I am struggling on the questions in the exercise.
Happy new year to you too! OK so the SUVAT equations work with positive and negatives for displacement, velocity and acceleration, as these are all vector quantities. What I mean by this, is if one velocity is positive and one is negative, they are going in different directions. When you choose your values for SUVAT, you are assigning them a direction, but there is no agreed positive direction, so we decide that. It doesn’t matter which direction you choose as positive, as long as you stay consistent with it for your calculations. You should try doing the same problem twice, with different positive directions, and you’ll end up with the same conclusion! I hope that helps!
The acceleration is always acting downwards. The velocity changes from positive to negative or vice versa, and the + or - takes into account the direction - it doesn't matter whether the acceleration is working with or against it, it is always downwards!
@@BicenMathsi had the same question. Just logically thinking about this, if you throw a ball in the air. The first half will have gravity opposing it and acceleration will be against it. The second half will have gravity supporting the ball’s fall so it should be quicker. Also surely it depends on how much force a person throws the ball - do those questions come up in mechanics?
@@BBK583 These things are all taken into account with the formulae and the positivity/negativity of the values. The force it is given doesn’t matter, we aren’t considering forces, only velocities - and these are separate entities.
In order to find the greatest height reached by the ball, would it not be easier to model it as a quadratic curve using s=ut+0.5at^2 , and then find the y value of the turning point (using calculator)? And then add this to the height it was projected from? Thanks for the videos they are great!
This absolutely would be an option, but I personally prefer this method, and it’s the traditional approach. However, if your method is clear and you end up with the same answer, you’ll always score the full marks (unless it has explicitly asked for a particular method of course!). And you’re welcome, pleased the videos are helpful!
It makes no difference, but I usually say the direction it is moving in is the positive - I tend to find this makes more sense in my head, but either way is perfectly fine!
At 18:55 you say u is -21 because you go upwards but I thought in displacement wise, you just only look at start and end so it’s actually going downwards
@@BicenMaths so at first speed goes up and so “u” would be negative but then the final velocity (when it lands), that is going down of course so would v be positive?
i dont understand the concept of '' if up was positive then a is -9.8 ......'' i use the the fact that when an object is moving up its decelerating so a=-9.8 and when its moving down the object is accelerating so a is +9.8. please let me know which concept to use.
Hi sir, how do we know that the distance is down or up at 7:08? Thanks a lot. Is it because the acceleration must be in the same direction as the distance?
So you always set a direction for the positive values. We set the direction for positive as upwards, which meant acceleration is negative as it downwards. As the particle is displaced downwards by 1.4m (as in it finishes up 1.4m below where it started), that also takes a negative value. You could easily try the question again by saying that down is positive instead - this would flip all the signs of the things in SUVAT, except for time (as time is not a vector quantity). You’ll still end up with the same answer!
