The distance s described by a particle in t seconds is given by s-aet-bet- Then the acceleration of the particle at time t is equal to

It is given that, the nbsp;distance nbsp;s nbsp;described by a particle in t seconds is given by nbsp;s=aet+bet. The acceleration of the particle at t ...

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Standard VII

Physics

Deceleration/Retardation

The distance ...

Question

The distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ . Then the acceleration of the particle at time t is equal to _____________.

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Solution

It is given that, the distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ .

The acceleration of the particle at time t is given by $\frac{d^{2} s}{d t^{2}}$ .

$s = a e^{t} + \frac{b}{e^{t}}$

Differentiating both sides with respect to t, we get

$\frac{d s}{d t} = \frac{d}{d t} (a e^{t} + \frac{b}{e^{t}})$

$\Rightarrow \frac{d s}{d t} = \frac{d}{d t} (a e^{t} + b e^{- t})$

$\Rightarrow \frac{d s}{d t} = a e^{t} + b e^{- t} \times (- 1)$

$\Rightarrow \frac{d s}{d t} = a e^{t} - b e^{- t}$

Again differentiating both sides with respect to t, we get

$\frac{d}{d t} (\frac{d s}{d t}) = \frac{d}{d t} (a e^{t} - b e^{- t})$

$\Rightarrow \frac{d^{2} s}{d t^{2}} = a e^{t} - b e^{- t} \times (- 1)$

$\Rightarrow \frac{d^{2} s}{d t^{2}} = a e^{t} + \frac{b}{e^{t}}$ = s

Thus, the acceleration of the particle at time t is $a e^{t} + \frac{b}{e^{t}}$ .

The distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ . Then the acceleration of the particle at time t is equal to $a e^{t} + \frac{b}{e^{t}}$ .

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The distance s described by a particle in t seconds is given by s=aet+bet. Then the acceleration of the particle at time t is equal to _____________.

The distance s described by a particle in t seconds is given by $s = a e^{t} + \frac{b}{e^{t}}$ . Then the acceleration of the particle at time t is equal to _____________.