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.

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- ParthKohli

- katieb

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- ParthKohli

@.Sam.

- anonymous

The highest velocity.

- ParthKohli

Note that the known are \(\alpha, \beta, T\).

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## More answers

- .Sam.

|dw:1368898722315:dw|

- ParthKohli

It's a particle, not a cube :-P

- .Sam.

Is the whole journey takes time T?

- ParthKohli

yes.

- .Sam.

|dw:1368898894389:dw|

- .Sam.

\[\alpha=\frac{v_1-0}{t_1} \]
Highest velocity is \(v_1\)
\[v_1=\alpha t_1\]

- .Sam.

\[-\beta=\frac{v_2-v_1}{t_2}\]
and also
\[T=t_1+t_2\]

- ParthKohli

Good going.

- .Sam.

Average velocity will be just \[\frac{v_1+v_2}{2}?\]

- ParthKohli

Careful: you don't know the average acceleration as yet.

- .Sam.

\[a_{ave}=\frac{v_2-v_1}{t_2-t_1}\]

- ParthKohli

Actually, the average acceleration for the whole thing is \(0\).

- .Sam.

Hmm, lol, but it didn't say it stops at the end and usually when its average acceleration \(\alpha=\beta\)

- ParthKohli

Oh, my bad. It does stop at the end.

- .Sam.

Hmm \[v_2=0\]

- ParthKohli

What is \(v_1\)? The maximum velocity or the start velocity?

- .Sam.

The diagram said it's maximum velocity

- ParthKohli

OK. I got confused when you said v1 = v2

- .Sam.

Nah

- .Sam.

So we'll mix them up
\[v_1=\alpha t_1 ~~~~~~ -\beta t_2=-v_1 \]
Getting
\[\beta t_2= \alpha t_1\]

- ParthKohli

You just eliminated \(v_1\) when you actually need it.

- .Sam.

Right,
\[v_1=\frac{\alpha(T-t_2)}{\beta}\]

- ParthKohli

But \(t_2\) is not a known thing. You defined it. Express in terms of alpha, beta and T.

- .Sam.

What about the distance?

- .Sam.

/displacement

- ParthKohli

You have to calculate that. Both 1 and 2 are to be given in terms of alpha, beta and T.

- ParthKohli

Do you want to try more?

- .Sam.

I'm getting more variables, lol

- .Sam.

But it would've been easier if we used calculus method

- ParthKohli

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 \(v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }\)

- .Sam.

Aw man it was getting close :P

- ParthKohli

:-D

- ParthKohli

Try the 2nd one maybe?

- .Sam.

Average velocity for the whole journey is it?

- ParthKohli

Yeah.

- .Sam.

is it zero

- ParthKohli

No, because the displacement isn't zero.

- .Sam.

But the initial and final velocities are zero

- ParthKohli

I have not asked for the average acceleration.

- ParthKohli

:-P

- ParthKohli

\[\overline{v} = \dfrac{\Delta s}{\Delta t}\]

- .Sam.

I've got
\[\frac{x_2}{x_3-x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}\]

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