IIT JEE , AIEEE Forum - Physics , Mechanics

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A particle moves with deceleration along a circle of radius so that at any moment its tangential and normal acc. are equal in magnitude.At initial moment t=0 ,the speed of the particle equal to v

Q1.Speed of the particle as a function of time t is given by v(t)=v/k1+k2*v*t/k3*R

thevyzz

normal acc = v^2/r

tangentrial acc = normal acc = v^2/r

considering v purley as speed

-dv/dt = tangential acc = v^2/r

-dv/v^2 = dt/r

integrating

+1/v = t/r + C

at t=0 v = v(intial)

+1/v(initial) = C

so+1/v = t/r +1/v(initial)

so,

v = (Vo*t+r )/ Vo*r ( Vo - intial velocity)

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