IIT JEE , AIEEE Forum - Physics , Mechanics

avik19

in the second law of kinematic eqn

x = ut + 1/2 at*2

if u = 0 then for constant acceleration t =

ca anyone xplain me the double answer????????

pink_ele

netagive tym corresponds to d case-when body cumes to rest after travelling distance x(which will be -ve) n deacceleration(-ve) ...........here at dis x...tym will be past one(-ve)..........
second ......+ve case is general case.............

tarinbansal

I dont think here negative time has some logical meaning.
I think its just becoz its a quadratic and u get 2 roots of a quadratic.

Take, x = 2 m, a = 1 m/s^2.
U get, t = +,- 2 sec

I dont think t = -2 sec has some significance.

PLZ correct me if I am wrong.

asish

@ tarin
it has a meaning physically
it shows the time b4 the observation started whwer u bcame 0
hope u get it

krishna.gopal

It means that it was at same position some time before as well If that motion is uniform acceleration from -sqrt(2x/a) to sqrt92x/a)

avik19

still not clear guys...........

does it mean that the negative sign has no meaning physically ..........but graphically it has some..............????????

plzzz correct me if i m wrong .................

pantpranav

Negative time is meaningless.

snail

I think it is like this :

Consider a stone falling from a height.

h=20m

a=10/m^2

u=0

v=u+at

x=1/2at^2

t= +/- root(2x/a)

t= 2 s ,-2 s

v=0+10 t

v=20m/s,-20m/s

Here you know that after 2s the stone will reach the ground traveling 20 m.

If you go back to -2 s , in this gravity system we know that the stone should have been thrown up 2 seconds before with velocity=-20m/s(ie. upwards) Then only it can reach the starting point at time=0 , u=0 ,h=20m.

In other words the stone was at x=20m even when t=-2 s provided we did not physically carry the stone to the height of 20 m and dropped it from there.

I hope it is clear!

feynmann

See, when u make an observation u do the zero-setting of time completely arbitrarily (i.e. as per ur convenenience).

So the same physical phenomena may be going on before ur specified zero - time ,just as well !!

what is 'u' ? it is the velocity at that arbitary zero-time . u= 0 means that u count ur time when the velocity is zero .

at t= - sqrt(2x/a) before ur specified time -zero the paticle was at same position after t= sqrt(2x/a) provided the particle was accelarting just the same way as before and after the time-zero .

So it is completely feasible ( -ve time ) . But if it is given in the qn that the particle was at rest before and then started accelarating after t>0 , then it is simply violating the underlined condition and it is not giving the correct answer .

I hope it is clear now .

sahana_2193

we usually reject the negetive part of the answer we get.but its not true.IT HAS A PHYSICAL MEANING. if u see any graph of motion we take the negetive tym.That depends upon where v start from...lyk the tym at which we start observing the motion...we frame it as t=0s,but even before our observation the body must be in a state of rest or motion whatever it is.

sahana_2193

to indicate that tym that is before our observation we take the tym negetive,But lemme tell ya it ha(-ve tym)more application in graphs.In numericals we usually reject the negetive case.hope my answer helped u

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