The distance s (in meters) covered by a parti | vedclass

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(C) Given the distance function: $s = ae^t + \frac{b}{e^t}$.To find the velocity $v$,we differentiate $s$ with respect to $t$:$v = \frac{ds}{dt} = \fr

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$s$

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(C) Given the distance function: $s = ae^t + \frac{b}{e^t}$.To find the velocity $v$,we differentiate $s$ with respect to $t$:$v = \frac{ds}{dt} = \frac{d}{dt}(ae^t + be^{-t}) = ae^t - be^{-t}$.To find the acceleration $a_{acc}$,we differentiate the velocity $v$ with respect to $t$:$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(ae^t - be^{-t}) = ae^t - b(-e^{-t}) = ae^t + be^{-t}$.Comparing this result with the original expression for $s$,we see that $a_{acc} = s$.Therefore,the acceleration is equal to the distance $s$.

The distance $s$ (in meters) covered by a particle in $t$ seconds is given by $s = ae^t + \frac{b}{e^t}$. Then the acceleration of the particle at time $t$ is:

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