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OK, here's a nice question that I was just checking out. Try it :-D A particle is at rest and starts to move at a constant acceleration \(\alpha\). After some time, it starts decelerating at \(\beta\) and returns to rest. Given that the particle takes time \(T\) to do so, find in the terms of the known: 1: The highest velocity. 2: The average velocity.

Physics
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@.Sam.
The highest velocity.
Note that the known are \(\alpha, \beta, T\).

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Other answers:

|dw:1368898722315:dw|
It's a particle, not a cube :-P
Is the whole journey takes time T?
yes.
|dw:1368898894389:dw|
\[\alpha=\frac{v_1-0}{t_1} \] Highest velocity is \(v_1\) \[v_1=\alpha t_1\]
\[-\beta=\frac{v_2-v_1}{t_2}\] and also \[T=t_1+t_2\]
Good going.
Average velocity will be just \[\frac{v_1+v_2}{2}?\]
Careful: you don't know the average acceleration as yet.
\[a_{ave}=\frac{v_2-v_1}{t_2-t_1}\]
Actually, the average acceleration for the whole thing is \(0\).
Hmm, lol, but it didn't say it stops at the end and usually when its average acceleration \(\alpha=\beta\)
Oh, my bad. It does stop at the end.
Hmm \[v_2=0\]
What is \(v_1\)? The maximum velocity or the start velocity?
The diagram said it's maximum velocity
OK. I got confused when you said v1 = v2
Nah
So we'll mix them up \[v_1=\alpha t_1 ~~~~~~ -\beta t_2=-v_1 \] Getting \[\beta t_2= \alpha t_1\]
You just eliminated \(v_1\) when you actually need it.
Right, \[v_1=\frac{\alpha(T-t_2)}{\beta}\]
But \(t_2\) is not a known thing. You defined it. Express in terms of alpha, beta and T.
What about the distance?
/displacement
You have to calculate that. Both 1 and 2 are to be given in terms of alpha, beta and T.
Do you want to try more?
I'm getting more variables, lol
But it would've been easier if we used calculus method
OK, so like \(v_1 = \alpha t_1 \Rightarrow t_1 = \dfrac{v_1}{\alpha}\). Similarly \(0 = v_1 - \beta t_2 \Rightarrow t_2 = \dfrac{v_1}{\beta}\). Now\[T = t_1 + t_2 = \frac{v_1}{\alpha} + \frac{v_1}{\beta}\]Solving for \(v_1 \), we get \(v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }\)
Aw man it was getting close :P
:-D
Try the 2nd one maybe?
Average velocity for the whole journey is it?
Yeah.
is it zero
No, because the displacement isn't zero.
But the initial and final velocities are zero
I have not asked for the average acceleration.
:-P
\[\overline{v} = \dfrac{\Delta s}{\Delta t}\]
I've got \[\frac{x_2}{x_3-x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}\]

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