Tricky Points on s-t and v-t Graphs - A Level Physics

In this video I recap a few of the essential things you need to understand about displacement-time and velocity-time graphs, including the significance of the gradient and area.
Thanks for watching,
Lewis
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[@stephenabraham2448]

im not doing a level maths along with physics and it makes everything harder since i got put in applications or something for not trying in my first 3 years of school and i wanted to take maths and physics at uni. so these videos really help. im taking a level maths next year though

[@murdermelon6103]

Since at one point he has 0 velocity wouldn’t it go down so a flat line at the bottom ?

[@tariyaposhar8363]

Thank you for the great explanation! Just a query, when the displacement-time graph has a curve does that show acceleration as well as instantaneous velocity?

[@alextmss]

Yes I think so :) if the line is straight there is a constant velocity but if the line curves then there must be a changing velocity (i.e positive or negative acceleration)

[@themediumcheese7484]

I don't fully understand the purpose of the tangent at 2:00.
Why draw the line at all if you can just get the values of ds and dt and punch them into the v-inst equation?
What difference does the tangent make when they're kind of arbitrarily drawn onto the graph?

[@PhysicsOnline]

the medium cheese If you have a constant velocity then you can use the gradient to calculate this. With the graph above you have curved lines and a non-uniform velocity. That means the only way to calculate it is to draw a tangent to the line at that point and calculate the gradient of this line. Otherwise you'd only be finding the average velocity.

[@themediumcheese7484]

Thank you for the quick reply!
I went ahead in your videos figuring that you'd touch back on this and it would click for me. And it did in your Deriving the Equations of Motion video. Cheers!
In essence, drawing a tangent and drawing an average line are basically the same technique applied in different ways - getting a straight line so we have a "constant" which we can lable as acceleration?
EDIT: Realised it was a bit vague. But basically determining the instantaneous velocity is just working out the average velocity for a very, very small instance of time right?

[@PhysicsOnline]

the medium cheese Hopefully as you see more videos you see the same ideas explained again and in slightly different ways.

[@user-xw4mu6nz4t]

So I noticed in earlier videos, he used x for displacement and now he uses s. Is that because the videos are just old and the exam boards have changed which symbols are used?

[@aleezabukhsh1878]

yeah ig because u now use "s" for displacement according to different sources but idrk

[@ritwikgossain]

What if time is negative?