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# A particle moves with deceleration along the circle of radius R so that at any moment of time its tangentialand normal accelerations are equal in moduli. At the initial moment t = 0 the speed of the particle equals v0,then :(i) the speed of the particle as a function of the distance covered s will be(A) v = v0 e–s/ R(B) v = v0 es/ R(C) v = v0 e–R/s (D) v = v0 eR/s

## 1 Answers

2 years ago
Dear Soham

at any instant tangential acceleraion is given by = dV/dt

but this is equal to centripital acceleation which is given as = mV2/r

so   dV/dt = V2/r

dV/V2 = 1/r dt

∫dV/V2 =∫ 1/r dt

-1/V = 1t/r + C

apply condition at t=0 V=u

so C=-1/c

equation become   -1/V = 1t/r + -1/u

1/V = 1/u  + t/r

1/V =(r+tu)/ur

V=ur/(r+tu)

so clearly after infinite time particle will come to rest

Regards

Arun (askIITians forum expert)

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