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??
options are root s/a root 2s/a 1/2roots s/a 2 root s/a
option c is correct but can anyone solve this
ok. Let me see.
ok plzzz go on...
What's the formula for kinetic energy?
Ek=mv^2/2
hey we have to solve this my kinematics equation (motion)
And, what's the formula for accleration? v=at right?
v=u+at
\(\huge E_k=\frac{mv^2}{2}\)
But, \(\huge v=at\)
distance = rate*time --> t = d/r rate = velocity = a*t --> t = s/at --> t^2 = s/a --> t = sqrt(s/a)
So. \(\huge E_k=\frac{m(at)^2}{2}\)
\(E_k\) is a constant.
So is m.
plzzz guys use kinemaics equation
Which one?
\(\Delta x=V_0t+\frac{1}{2}at^2\)?
kinematic equations...this is why i stick with just math...i showed you the answer
Wait, I got it. I'm so silly.
\(s=V_0t+\frac{1}{2}at^2\) Remember how to find the vertex of a quadtratic equation?
i think we have to use s=ut+1/2 at^2
yes plzz go on inkyvoyd
Now. the vertex of a quadtratic is at \(\huge (-\frac{1}{2a},-\frac{D}{4a})\) where \(D=\sqrt{b^2-4ac}\)
then
can u give the answer completely... plzz
Eh, I don't hav eit, @shameer1
i cant understand what u have written
I can't either.
oh....ohh....no
Let me think :S
I'm not sure I understand the problem
Can you restate it?
what???
"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?
"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
oooh
ok leave this i will post another question
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