The displacement `x` of a particle varies with time `t` as `x = ae^(-alpha t) + be^(beta t)`. Where `a,b, alpha` and `beta` positive constant.
The velocity of the particle will.

Options

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.

Explanation

To solve the problem, we need to find the velocity of a particle whose displacement \( x \) varies with time \( t \) as given by the equation:

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

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

### Step 1: Find the velocity \( v \)

The velocity \( v \) of the particle is the first derivative of the displacement \( x \) with respect to time \( t \):

\[ v = \frac{dx}{dt} \]

Differentiating the expression for \( x \):

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

Using the chain rule for differentiation, we get:

\[ v = a \cdot \frac{d}{dt}(e^{-\alpha t}) + b \cdot \frac{d}{dt}(e^{\beta t}) \]

Calculating the derivatives:

\[ \frac{d}{dt}(e^{-\alpha t}) = -\alpha e^{-\alpha t} \]
\[ \frac{d}{dt}(e^{\beta t}) = \beta e^{\beta t} \]

Substituting these back into the equation for \( v \):

\[ v = a(-\alpha e^{-\alpha t}) + b(\beta e^{\beta t}) \]

Thus, we have:

\[ v = -\alpha ae^{-\alpha t} + b\beta e^{\beta t} \]

### Step 2: Find the acceleration \( a \)

The acceleration \( a \) of the particle is the derivative of the velocity \( v \) with respect to time \( t \):

\[ a = \frac{dv}{dt} \]

Differentiating the expression for \( v \):

\[ a = \frac{d}{dt}(-\alpha ae^{-\alpha t} + b\beta e^{\beta t}) \]

Using the chain rule again:

\[ a = -\alpha a \cdot \frac{d}{dt}(e^{-\alpha t}) + b\beta \cdot \frac{d}{dt}(e^{\beta t}) \]

Calculating the derivatives again:

\[ \frac{d}{dt}(e^{-\alpha t}) = -\alpha e^{-\alpha t} \]
\[ \frac{d}{dt}(e^{\beta t}) = \beta e^{\beta t} \]

Substituting these back into the equation for \( a \):

\[ a = -\alpha a(-\alpha e^{-\alpha t}) + b\beta(\beta e^{\beta t}) \]

This simplifies to:

\[ a = \alpha^2 ae^{-\alpha t} + b\beta^2 e^{\beta t} \]

### Step 3: Analyze the acceleration

Since \( a, b, \alpha, \) and \( \beta \) are all positive constants, both terms \( \alpha^2 ae^{-\alpha t} \) and \( b\beta^2 e^{\beta t} \) are positive for all \( t \). Therefore, the acceleration \( a \) is always greater than zero:

\[ a > 0 \quad \text{for all } t \]

### Step 4: Conclusion about the velocity

Since the acceleration is positive, this implies that the velocity \( v \) is increasing over time. Therefore, we can conclude that:

- The velocity of the particle will be increasing.

### Final Answer:
The velocity of the particle will be increasing.

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