- #1
konichiwa2x
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The displacement 'x' of a particle varies with time according to the relation
[tex] x = \frac{a}{b}(1 - e^-^b^t)[/tex] Then,
(A) At t=1/b the displacement of the particle is a/b.
(B) The velocity and acceleration of the particle at t=0 are 'a' and 'b' respectively.
(C) The particle cannot reach a point at a distance 'x' from its starting point if x'>ab.
(D) The particle will come back to its starting point as 't' tends to infinity.
I have differentiated to obtain the velocity and also acceleration but I still can't seem to solve this.
My book says the correct answer is (C) but how do you arrive at it? Please help.
[tex] x = \frac{a}{b}(1 - e^-^b^t)[/tex] Then,
(A) At t=1/b the displacement of the particle is a/b.
(B) The velocity and acceleration of the particle at t=0 are 'a' and 'b' respectively.
(C) The particle cannot reach a point at a distance 'x' from its starting point if x'>ab.
(D) The particle will come back to its starting point as 't' tends to infinity.
I have differentiated to obtain the velocity and also acceleration but I still can't seem to solve this.
My book says the correct answer is (C) but how do you arrive at it? Please help.