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OpenStudy (anonymous):

the greatest acceleration or deceleration that a train may have is a the minimum time in which the train can start from one station to stop at the next ( at a straight distance of s ) is??

OpenStudy (anonymous):

options are root s/a root 2s/a 1/2roots s/a 2 root s/a

OpenStudy (anonymous):

option c is correct but can anyone solve this

OpenStudy (inkyvoyd):

ok. Let me see.

OpenStudy (anonymous):

ok plzzz go on...

OpenStudy (inkyvoyd):

What's the formula for kinetic energy?

OpenStudy (anonymous):

Ek=mv^2/2

OpenStudy (anonymous):

hey we have to solve this my kinematics equation (motion)

OpenStudy (inkyvoyd):

And, what's the formula for accleration? v=at right?

OpenStudy (anonymous):

v=u+at

OpenStudy (inkyvoyd):

\(\huge E_k=\frac{mv^2}{2}\)

OpenStudy (inkyvoyd):

But, \(\huge v=at\)

OpenStudy (dumbcow):

distance = rate*time --> t = d/r rate = velocity = a*t --> t = s/at --> t^2 = s/a --> t = sqrt(s/a)

OpenStudy (inkyvoyd):

So. \(\huge E_k=\frac{m(at)^2}{2}\)

OpenStudy (inkyvoyd):

\(E_k\) is a constant.

OpenStudy (inkyvoyd):

So is m.

OpenStudy (anonymous):

plzzz guys use kinemaics equation

OpenStudy (inkyvoyd):

Which one?

OpenStudy (inkyvoyd):

\(\Delta x=V_0t+\frac{1}{2}at^2\)?

OpenStudy (dumbcow):

kinematic equations...this is why i stick with just math...i showed you the answer

OpenStudy (inkyvoyd):

Wait, I got it. I'm so silly.

OpenStudy (inkyvoyd):

\(s=V_0t+\frac{1}{2}at^2\) Remember how to find the vertex of a quadtratic equation?

OpenStudy (anonymous):

i think we have to use s=ut+1/2 at^2

OpenStudy (anonymous):

yes plzz go on inkyvoyd

OpenStudy (inkyvoyd):

Now. the vertex of a quadtratic is at \(\huge (-\frac{1}{2a},-\frac{D}{4a})\) where \(D=\sqrt{b^2-4ac}\)

OpenStudy (anonymous):

then

OpenStudy (anonymous):

can u give the answer completely... plzz

OpenStudy (inkyvoyd):

Eh, I don't hav eit, @shameer1

OpenStudy (anonymous):

i cant understand what u have written

OpenStudy (inkyvoyd):

I can't either.

OpenStudy (anonymous):

oh....ohh....no

OpenStudy (inkyvoyd):

Let me think :S

OpenStudy (inkyvoyd):

I'm not sure I understand the problem

OpenStudy (inkyvoyd):

Can you restate it?

OpenStudy (anonymous):

what???

OpenStudy (inkyvoyd):

"the greatest acceleration or deceleration that a train may have is a the minimum time in which the train can start from one station to stop at the next ( at a straight distance of s ) is?? what does "may have is a the minimum time in which the train can start from one station to stop at the next" mean?

OpenStudy (anonymous):

"the greatest acceleration or deceleration that a train may have is a. the minimum time in which the train can start from one station to stop at the next ( at a straight distance of s

OpenStudy (inkyvoyd):

oooh

OpenStudy (anonymous):

ok leave this i will post another question

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