Question

The displacement x of a particle varies with time t as \(x=ae^{-  \alpha t} + be ^{\beta t}\), where \(a,b,\) \(\alpha\) and \(\beta\) are positive constants. The velocity of the particle will:

• A. go on decreasing with time
• B. be independent of \(\alpha\) and \(\beta\)
• C. drop to zero when \(\alpha\) = \(\beta\)
• D. go on increasing with time

Answer 1

The Correct Option is D

To determine how the velocity of the particle changes with time, we need to first find the velocity expression from the given displacement equation:

x = ae^{-\alpha t} + be^{\beta t}

Velocity is the derivative of displacement with respect to time, denoted as v = \frac{dx}{dt}.

Calculating the derivative, we have:

\(\frac{dx}{dt} = \frac{d}{dt}(ae^{-\alpha t} + be^{\beta t})\)

• For ae^{-\alpha t}, the derivative is -\alpha ae^{-\alpha t}.
• For be^{\beta t}, the derivative is \beta be^{\beta t}.

Thus, the velocity v is given by:

v = -\alpha ae^{-\alpha t} + \beta be^{\beta t}

Examining the components of this equation:

• The term -\alpha ae^{-\alpha t} is always negative and decreases with time because e^{-\alpha t} decreases as t increases.
• The term \beta be^{\beta t} is always positive and increases with time because e^{\beta t} increases as t increases.

Therefore, the positive, increasing component \beta be^{\beta t} will dominate the velocity behavior as time progresses. Hence, the velocity of the particle will go on increasing with time.

Based on the analysis, the correct answer is that the velocity will go on increasing with time.