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. decrease with time
B. be independent of `alpha` and `beta`
C. drop to zero when `alpha = beta`
D. increase with time

Answer

To find the velocity of the particle whose displacement varies with time as given by the equation \( x = ae^{-\alpha t} + be^{\beta t} \), we will follow these steps:

### Step 1: Write down the displacement equation
The displacement \( x \) of the particle is given by:
\[ x(t) = ae^{-\alpha t} + be^{\beta t} \]
where \( a, b, \alpha, \) and \( \beta \) are positive constants.

### Step 2: Differentiate the displacement to find velocity
The velocity \( v \) of the particle is the time derivative of displacement \( x \):
\[ v(t) = \frac{dx}{dt} \]
Using the chain rule, we differentiate \( x(t) \):
\[ v(t) = \frac{d}{dt}(ae^{-\alpha t} + be^{\beta t}) \]

### Step 3: Apply the differentiation
Now, we differentiate each term:
1. For the first term \( ae^{-\alpha t} \):
\[ \frac{d}{dt}(ae^{-\alpha t}) = a \cdot (-\alpha) e^{-\alpha t} = -a\alpha e^{-\alpha t} \]
2. For the second term \( be^{\beta t} \):
\[ \frac{d}{dt}(be^{\beta t}) = b \cdot \beta e^{\beta t} = b\beta e^{\beta t} \]

Combining these results, we get:
\[ v(t) = -a\alpha e^{-\alpha t} + b\beta e^{\beta t} \]

### Step 4: Analyze the velocity equation
The velocity equation is:
\[ v(t) = -a\alpha e^{-\alpha t} + b\beta e^{\beta t} \]
Since \( e^{-\alpha t} \) and \( e^{\beta t} \) are both positive for all \( t \), the signs of the terms depend on the constants \( a, \alpha, b, \) and \( \beta \).

### Step 5: Determine the behavior of velocity
To analyze whether the velocity increases or decreases with time, we need to find the acceleration \( a(t) \) by differentiating the velocity:
\[ a(t) = \frac{dv}{dt} \]
Differentiating \( v(t) \):
\[ a(t) = \frac{d}{dt}(-a\alpha e^{-\alpha t} + b\beta e^{\beta t}) \]
This gives:
1. For the first term:
\[ \frac{d}{dt}(-a\alpha e^{-\alpha t}) = a\alpha^2 e^{-\alpha t} \]
2. For the second term:
\[ \frac{d}{dt}(b\beta e^{\beta t}) = b\beta^2 e^{\beta t} \]

Combining these, we find:
\[ a(t) = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t} \]

### Step 6: Conclusion about velocity
Since \( a, b, \alpha, \) and \( \beta \) are all positive constants, both terms in \( a(t) \) are positive:
\[ a(t) > 0 \quad \text{for all } t \]
This means that the acceleration is always positive, indicating that the velocity \( v(t) \) is increasing with time.

### Final Answer
The velocity of the particle will **increase with respect to time**.

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