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

Correct answer: D. go on increasing with time Given, x=a e^ -α t +b e^ β t So, velocity aligned v & =(d x)/(d t) & =-a α e^ -α t +b β e^ β t & =A+B aligned Where, A=-a α e^ -α t , B=b β e^ β t The value of term A=-a α e^ -α t decreases and of term B=b β e^ β t increases with increase in time. As a result, velocity goes on increasing with time.

JIPMER 2007

Exam: JIPMER
Year: 2007
Subject: Physics
Chapter: Motion in One Dimension

JIPMER 2007 · Physics · Motion in One Dimension

The displacement x of a particle varies with time t as x=a e^ -α t...

The displacement

x

of a particle varies with time

t

x=a e^{-\alpha t}+b e^{\beta t}

, where

a, b, \alpha

and

\beta

are positive constants. The velocity of the particle will

Options

Ago on decreasing with time
Bbe independent of $\alpha$ and $\beta$
Cdrop to zero when $\alpha=\beta$
Dgo on increasing with time

Correct answer

D. go on increasing with time

Step-by-step solution

Given,

x=a e^{-\alpha t}+b e^{\beta t}

So, velocity

\begin{aligned} v & =\frac{d x}{d t} \\ & =-a \alpha e^{-\alpha t}+b \beta e^{\beta t} \\ & =A+B\end{aligned}

Where,

A=-a \alpha e^{-\alpha t}, B=b \beta e^{\beta t}

The value of term

A=-a \alpha e^{-\alpha t}

decreases and of term

B=b \beta e^{\beta t}

increases with increase in time. As a result, velocity goes on increasing with time.

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