IIT JEE , AIEEE Forum - Physics , Mechanics

kislay

pls solve

nishant

we know that a=dv/dt..from the relation between acceleration and velocity given,we get a=4/v ms^-2....now,taking v to left side and bringing dt to right side,we have v dv = 4 dt..this gives v^2/2 =4t.now,integrating v dv from 6m/s to v m/s and dt from 2s to 3s,we get v^2/2-18=4*3-4*2 which implies v^2/2=22m/s giving v as

44 or 6.63 m/s..acceleration is therefore,4/6.63=0.6 ms^-2..hope u got it..plz rate my answer if u find it satisfactory.don't hesitate to ask if u haven't understood anything in the solution...cheers!!

iitkgp_bipin

v = 4/a

va = 4

(dx/dt).(vdv/dx) = 4

vdv = 4dt

v²/2 = 4t + c

At t = 2 v = 6   gives   6²/2 = 4x2 + c    gives   c = 10

Hence    v² = 8t + 20

At t = 3    v² = 8x3 + 20 = 44

Hence a = 4/v = 4/

44 = 2/

11 m/s²

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