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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.
 11 months ago
 11 months ago
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.
 11 months ago
 11 months ago

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myiah5766Best ResponseYou've already chosen the best response.0
The highest velocity.
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Note that the known are \(\alpha, \beta, T\).
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
It's a particle, not a cube :P
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
Is the whole journey takes time T?
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
\[\alpha=\frac{v_10}{t_1} \] Highest velocity is \(v_1\) \[v_1=\alpha t_1\]
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
\[\beta=\frac{v_2v_1}{t_2}\] and also \[T=t_1+t_2\]
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
Average velocity will be just \[\frac{v_1+v_2}{2}?\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Careful: you don't know the average acceleration as yet.
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
\[a_{ave}=\frac{v_2v_1}{t_2t_1}\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Actually, the average acceleration for the whole thing is \(0\).
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
Hmm, lol, but it didn't say it stops at the end and usually when its average acceleration \(\alpha=\beta\)
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Oh, my bad. It does stop at the end.
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
What is \(v_1\)? The maximum velocity or the start velocity?
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
The diagram said it's maximum velocity
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
OK. I got confused when you said v1 = v2
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
So we'll mix them up \[v_1=\alpha t_1 ~~~~~~ \beta t_2=v_1 \] Getting \[\beta t_2= \alpha t_1\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
You just eliminated \(v_1\) when you actually need it.
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
Right, \[v_1=\frac{\alpha(Tt_2)}{\beta}\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
But \(t_2\) is not a known thing. You defined it. Express in terms of alpha, beta and T.
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
What about the distance?
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
You have to calculate that. Both 1 and 2 are to be given in terms of alpha, beta and T.
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Do you want to try more?
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
I'm getting more variables, lol
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
But it would've been easier if we used calculus method
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
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 \(v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }\)
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
Aw man it was getting close :P
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Try the 2nd one maybe?
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
Average velocity for the whole journey is it?
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
No, because the displacement isn't zero.
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
But the initial and final velocities are zero
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
I have not asked for the average acceleration.
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[\overline{v} = \dfrac{\Delta s}{\Delta t}\]
 11 months ago

.Sam.Best ResponseYou've already chosen the best response.2
I've got \[\frac{x_2}{x_3x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}\]
 11 months ago
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