The greatest acceleration ordeceleration that a train may have is a.find minimum time in which the train may reach from one station to other separated by a distance d
So what you want to do obviously is accelerate to the midpoint and decelerate from the mid-point to the end. Given that, how long does it take to reach the mid-point?
?
How far is the mid point from the first station? d/2. How far does the train travel if it starts at rest and accelerates at a constant rate of a?
c'mon Aravind. How far does a body move if it starts at rest and accelerates at a constant rate of a for a time of t?
If you're not going to stay and engage with me I don't know why I bother to try and help you. This isn't the first time you walk away mid-conversation and it gets pretty boring for me if you're not going to make an effort. Sorry.
srry james i will try not to repeat it
ok. So where did you get with this question?
Aravind, talk to me here, not in other people's questions.
k
Have you finished this question? Where are you with it?
i am doing another currently
i am discussing with turing can u also cm
well, I'd like to see you finish one question (with my help) before I engage with you on another. I'm unhappy that I've never seen you reach the end of a problem. When you're ready, let's finish this problem. Then I will help you with a new problem. But for now I'm not interested in helping you with a new one. When you're ready, let me know here.
K
for minimum time acceleration is a
james?
Oh, ok, you want to do this now. So the train accelerates to the half way point d/2, then decelerates.
The first question is, how long does it take to reach d/2 if it starts from rest. So if it accelerates at a constant rate, then velocity v at time t is v = at. What then is the displacement at time t?
If s(t) is the displacement, we want to know the time t such that s(t) = d/2. Hence we need a formula for s(t). What is that formula, given that v(t) = at ?
And you're not here? Where are you?
For the record \[ s(t) = s(0) + \int_0^t v(u) \ du \] \[ = s(0) + \int_0^t au \ du \] \[ = s(0) + \frac{1}{2} at^2 \] As s(0) = 0, hence \[ s(t) = \frac{1}{2}at^2 \]
ya
so 1/2 at^2 is the answer??
s(t) = 1/2 . at^2. So how long does it take the train to reach the mid-point d/2 ?
at^2
No. At the mid-point s(t) = d/2, as s(t) is the expression for displacement at time t. So the question is: for what value of t is this the case?
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