Displacement of a particle after time t is given by x = left( {k/{b^2}} right)left( {1 - {e^{ - bt}}

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2.Motion in Straight Line

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The displacement of a particle after time $t$ is given by $x = \left( {k/{b^2}} \right)\left( {1 - {e^{ - bt}}} \right)$ where $b$ is a constant. What is the acceleration of the particle?

A

$k{e^{ - bt}}$

B

$-k{e^{ - bt}}$

C

$\frac{k}{{{b^2}}}{e^{ - bt}}$

D

$\frac{-k}{{{b^2}}}{e^{ - bt}}$

Solution

$x=\frac{k}{b^{2}}\left(1-e^{-b t}\right)$

$\frac{d x}{d t}=\frac{k}{b} e^{-b t}, \frac{d^{2} x}{d t^{2}}=-k e^{-b t}$

Standard 11

Physics

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