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

Answer 1

The Correct Option is D

To solve the given problem, we need to understand the behavior of the velocity of a particle whose displacement is described by the equation:

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

where a, b, \alpha, and \beta are positive constants.

1. First, let's find the expression for velocity v. Velocity is the time derivative of displacement, so we differentiate x with respect to time t:

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

Using the chain rule, we have:

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

2. Now, let's analyze the behavior of the velocity over time:
3. The term -a\alpha e^{-\alpha t} exponentially decays to zero as time t increases since e^{-\alpha t} decreases.
4. The term b\beta e^{\beta t} exponentially increases with time since e^{\beta t} increases.
5. Therefore, as time progresses, the positive term b\beta e^{\beta t} will dominate the negative term, causing the overall velocity v to increase over time.

From this analysis, we can conclude that the velocity of the particle goes on increasing with time, thus justifying the correct answer:

go on increasing with time

The incorrect options can be ruled out as follows:

1. "be independent of \beta": This is incorrect because \beta directly affects the growth rate of the positive exponential term b\beta e^{\beta t}.
2. "drop to zero when \alpha = \beta": While velocities can cancel under specific conditions, in this scenario, the terms will not generally cancel at \alpha = \beta, especially since a and b could still be different.
3. "go on decreasing with time": This is incorrect as the velocity's positive term grows with time and will eventually dominate.