The displacement of a particle varies with time t - x - ae - t - be - t- where a - b - - and - are positive constants- The velocity of the particle willA- Go on decreasing with timeB- Drop to zero when -C- Go on increasing with timeD- Be independent of - and -

The correct option is C Go on increasing with timex=ae^ α t+be^β t Velocity v=dx/dt=d/dt(ae^ α t+be^β t ) =ae^ α t( α)+be^β t(β)= aα e^ α t+b β e^β t From t ...

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Byju's Answer

Standard IX

Physics

Speed / Velocity Time Graphs

The displacem...

Question

The displacement of a particle varies with time t, $x = a e^{- α t} + b e^{β t}$ , where a, b, $α$ and $β$ are positive constants. The velocity of the particle will

Go on decreasing with time

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Be independent of α and β

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Drop to zero when α = β

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Go on increasing with time

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Solution

The correct option is (C) Go on increasing with time
Given,
$x = a e^{- α t} + b e^{β t}$
We know,
Velocity $v = \frac{d x}{d t} = \frac{d}{d t} (a e^{- α t} + b e^{β t})$

$= a e^{- α t} (- α) + b e^{β t} (β) = - a α e^{- α t} + b β e^{β t}$

From the above equation itself we can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase

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The displacement of a particle varies with time t, x=ae−αt+beβt, where a, b, α and β are positive constants. The velocity of the particle will

The displacement of a particle varies with time t, $x = a e^{- α t} + b e^{β t}$ , where a, b, $α$ and $β$ are positive constants. The velocity of the particle will