IIT JEE , AIEEE Forum - Physics , Mechanics

tarun_bits

under constant acceleration

_

v = v1 + v2 / 2

is this still true if acceleration is not constant

amrita

i think it's not so.......

shubham_sachdeva

no its not true......

amrita

v=v1+v2/2 since

avg. velocity=total distance/total time

==> v=v1t+1/2at²/t = v1+1/2at = 2v1+at/2 = v1+v2/2.............[where v1 and v2 are initial and final velocities]

but in case of non uniform acceleration,

v=v1t1+1/2a1t1²+ v2t2+1/2a2t2² /(t1+t2)....................[v2 is the final velocity in the first condition and initial velocity in second condition]

v = t1(v1+v2) + t2(v2+v3)/ 2(t1+t2)

amrita

if u think m rite plzzzzz rate me

tarun_bits

good job amrita..its right

amrita

thanxxx

elessar_iitkgp

As we know for constant acceleration a,

x = v₀(

t) + (1/2)a(

t)²

x /

t= v₀ + (1/2)a(

t)

v_av = = v₀ + (1/2)a(

t)

v = a(

t)

t =

v/a
Substituting the value of

t in the second term of the first equation
v_av = v₀ + (1/2)a(

v/a) = v₀ + (1/2)(v - v₀)
v_av= (1/2)(v + v₀)
As can be seen this result is derived from the results for constant acceleration.
It isn't applicable to non uniform accelerations.

tarun_bits

stavan110

no i dont think it is so...

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