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alphabeta
 one year ago
How to find the instantaneous acceleration by looking at this graph (file attached) at time 0.50s?
alphabeta
 one year ago
How to find the instantaneous acceleration by looking at this graph (file attached) at time 0.50s?

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442065747164:dw Hey alphabeta! Acceleration is the rate of change in velocity or in other words, \(\sf acceleration = \Large \frac{dv}{dt}\)= Slope of Velocitytime graph. So, let's suppose we want to calculate the acceleration of the body at t=1 sec, then slope will roughly be equal to 6 over 1 = \(\sf \Large \frac{6}{1}\)

alphabeta
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @Abhisar ! That really helped a lot! :)

alphabeta
 one year ago
Best ResponseYou've already chosen the best response.0At least I thought I understood? For some reason, I don't get the right answer! I'm doing it like this: (4.2ms^1)/(0.50s) = 8.4 ms^2, but the markscheme says 4.5ms^2... But when thinking about it, isn't this the average acceleration rather than the instantaneous?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1What do you want to figure out, instantaneous acceleration or average acceleration?

alphabeta
 one year ago
Best ResponseYou've already chosen the best response.0Instantaneous (as written in my first post) :)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Well, in that case it should be 8.4 m/s^2. That's what I believe but let's call @IrishBoy123 to counter check it.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1442072505379:dwyou need the slope of the tangent line at t = 0.5 i think @Abhisar 's line is parallel and gives you a very good approximation so your denominator should be 1 in your calculation but it's just an approximation, isn't it :p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2that's for **instantaneous acceleration** at t = 0.5, to be clear. the slope at that instant

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, basically it needs to be a tangent drawn through the curve as you have drawn.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442092345797:dw I am approximating as we don't have a computer to be exact, and as you were still confused maybe this will clear things up a bit @alphabeta this is for instantaneous acceleration, keeping it simple \[a = \frac{ \Delta v }{ \Delta t }\] notice we take the slope of the line here at that instant.

alphabeta
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, now I see! Thanks soo much! :D
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