The displacement x of a particle varies with time according to the relation x=ab1−e−bt. Then take e−1≃13
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a
At t = 1/b, the displacement of the particle is nearly (1/3) (a/b)
b
The velocity and acceleration of the particle at t= 0 are a and ab respectively
c
The particle cannot reach a point at a distance x' from its starting position if x' > a/b
d
The particle will come back to its starting point as t→∞
answer is C.
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Detailed Solution
Velocity of the particle is given by v=dxdt=ddtab1−e−bt=ae−btAcceleration of the particle is given by α=dvdt=ddtae−bt=−abe−btAt t =1/b, the displacement of the particle is x=ab1−e−1≃ab1−13≃23ab ∵e−1≃13Hence choice (a) is wrong. At t= 0, the values v and α respectively are v=ae−0=a and α=−abe-0=-ab.Hence choice (b) is also wrong. The displacement x is maximum when t→∞,i.e., xmax=ab1-e-∞=ab.Hence choice (c ) is correct.