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A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and 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

A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential
and 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

Grade:11

1 Answers

Arun
25763 Points
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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