Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
ParthKohli
Group Title
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.
 one year ago
 one year ago
ParthKohli Group Title
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.
 one year ago
 one year ago

This Question is Closed

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

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
dw:1368898722315:dw
 one year ago

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

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
dw:1368898894389:dw
 one year ago

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

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Good going.
 one year ago

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

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

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

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

.Sam. Group TitleBest 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 \(\alpha=\beta\)
 one year ago

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
Hmm \[v_2=0\]
 one year ago

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

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

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

.Sam. Group TitleBest 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 \[\beta t_2= \alpha t_1\]
 one year ago

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

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

ParthKohli Group TitleBest 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.
 one year ago

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
/displacement
 one year ago

ParthKohli Group TitleBest 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.
 one year ago

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

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

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

ParthKohli Group TitleBest 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 \(v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }\)
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
Aw man it was getting close :P
 one year ago

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

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

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

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

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

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

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
I've got \[\frac{x_2}{x_3x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}\]
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.