The displacement x of a particle varies with time as `x=a e^(alphat)+be^(betat)` where a,b,a, `beta` are constants and are positives. The velocity of the particle will:

Options

A. Drop to zero when `alpha=beta`
B. be independnent of `alpha` and `beta`
C. go on increasng with time
D. go on descreasing 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:

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

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

### Step 1: Differentiate the displacement to find velocity
The velocity \( v \) of the particle is defined as the rate of change of displacement with respect to time. Therefore, we differentiate \( x \) with respect to \( t \):

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

### Step 2: Apply the differentiation
Using the chain rule for differentiation, we differentiate each term in the expression for \( x \):

1. Differentiate \( a e^{\alpha t} \):
- The derivative of \( e^{\alpha t} \) is \( \alpha e^{\alpha t} \).
- Therefore, the derivative of \( a e^{\alpha t} \) is \( a \alpha e^{\alpha t} \).

2. Differentiate \( b e^{\beta t} \):
- The derivative of \( e^{\beta t} \) is \( \beta e^{\beta t} \).
- Therefore, the derivative of \( b e^{\beta t} \) is \( b \beta e^{\beta t} \).

Combining these results, we get:

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

### Step 3: Analyze the velocity
Since \( a \), \( b \), \( \alpha \), and \( \beta \) are all positive constants, both terms \( a \alpha e^{\alpha t} \) and \( b \beta e^{\beta t} \) are positive for all \( t \). This means that:

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

### Step 4: Conclusion about the velocity
Since the velocity is a sum of two positive terms, the velocity of the particle will always be positive and will increase as time increases because both \( e^{\alpha t} \) and \( e^{\beta t} \) are exponential functions that grow with time.

### Final Answer
The velocity of the particle will always be positive and increasing.

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