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ParthKohli
 one year 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.
ParthKohli
 one year 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.

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myiah5766
 one year ago
Best ResponseYou've already chosen the best response.0The highest velocity.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Note that the known are \(\alpha, \beta, T\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1It's a particle, not a cube :P

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

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

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

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Careful: you don't know the average acceleration as yet.

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Actually, the average acceleration for the whole thing is \(0\).

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Oh, my bad. It does stop at the end.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1What is \(v_1\)? The maximum velocity or the start velocity?

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1OK. I got confused when you said v1 = v2

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1You just eliminated \(v_1\) when you actually need it.

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

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

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

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Do you want to try more?

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

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1OK, 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.
 one year ago
Best ResponseYou've already chosen the best response.2Aw man it was getting close :P

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Try the 2nd one maybe?

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1No, because the displacement isn't zero.

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I have not asked for the average acceleration.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\overline{v} = \dfrac{\Delta s}{\Delta t}\]

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