The distance in seconds- described by a particle in t seconds is given by s - ae t - b - e t- Then acceleration of the particle at time t isA- Proportional to tB- Proportional to sC- SD- Constant

The correct option is C sGiven that s=ae^t+b/e^t Differentiating w.r.t. time t, we get ds/dt =velocity =ae^t b/e^t Again d^2 s/dt^2 = acceleration =ae^t+b/e^t= ...

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Standard XII

Physics

Constrained Motion : Intuitive Approach

The distance ...

Question

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

Proportional to t

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Proportional to s

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Constant

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Solution

The correct option is C s
Given that $s = a e^{t} + \frac{b}{e^{t}}$
Differentiating w.r.t. time t, we get
$\frac{d s}{d t}$ =velocity $= a e^{t} - \frac{b}{e^{t}}$
Again $\frac{d^{2} s}{d t^{2}}$ = acceleration $= a e^{t} + \frac{b}{e^{t}} = s$

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

The correct option is C sGiven that s=aet+bet Differentiating w.r.t. time t, we get dsdt =velocity =aet−bet Again d2sdt2 = acceleration =aet+bet=s

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

The correct option is C s
Given that $s = a e^{t} + \frac{b}{e^{t}}$
Differentiating w.r.t. time t, we get
$\frac{d s}{d t}$ =velocity $= a e^{t} - \frac{b}{e^{t}}$
Again $\frac{d^{2} s}{d t^{2}}$ = acceleration $= a e^{t} + \frac{b}{e^{t}} = s$