25 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Where's the graph?
There is no graph, the graph is like part of the solution, that is why I got confused
it would matter right?
No, the graph is given according to the question.
yes
Basic premise of the solution is:
i.) Area of a-t graph gives change in velocity
ii.) If the function of acceleration is a=mx+c, then, $vdv=(mx+c)dx$, hence integrating, $v^{2}=mx^{2}+cx$
SirLancelotDuLac
So if you know the velocity at time t (easily deduced from graph) we can find position
Oh ok so the 6,7 on y axis is part of the question right?
Oh sorry, we have to find velocity...
the question says linear variation of a with time i think, not displacement along x axis...
Yep then the first point, just find area of a-t graph
yeah the acceleration time graph, even if specified as linear, would be unique for a case ☠️
Yes, that is what the first point covers.
Thank god I was like , how did it magically appear
which referance material is it from?
Cengage
How r they finding the area here, sry if this kinda dum
Area of trapezoid
Oh ok thankyou so much I forgot that existed
Got it, ty again
Welcome mate.
+solved @SirLancelotDuLac @Deleted User
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