The velocity-time relationship is described by equation `v = P + Qt^(2)`. The body is travelling with

Options

A. zero accalaration
B. uniform acceleration
C. nonuniform acceleration
D. None of the above

Explanation

To solve the problem, we need to analyze the given velocity-time relationship described by the equation:

\[ v = P + Qt^2 \]

where:
- \( v \) is the velocity,
- \( P \) is a constant,
- \( Q \) is a constant,
- \( t \) is the time.

### Step 1: Understand the relationship
The equation shows that the velocity \( v \) is a function of time \( t \). The term \( Qt^2 \) indicates that the velocity is changing with time.

### Step 2: Differentiate the velocity to find acceleration
To find the acceleration \( a \), we need to differentiate the velocity \( v \) with respect to time \( t \):

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

Differentiating \( v \):
\[ \frac{dv}{dt} = \frac{d}{dt}(P + Qt^2) \]

Since \( P \) is a constant, its derivative is 0. The derivative of \( Qt^2 \) is:

\[ \frac{d}{dt}(Qt^2) = 2Qt \]

Thus, the acceleration \( a \) becomes:

\[ a = 2Qt \]

### Step 3: Analyze the acceleration
Now, we have:

\[ a = 2Qt \]

This indicates that the acceleration depends on time \( t \). Since \( Q \) is a constant, the acceleration is not constant; it varies with time.

### Conclusion
Since the acceleration \( a = 2Qt \) changes with time, we conclude that the body is experiencing non-uniform acceleration.

### Final Answer
The body is travelling with **non-uniform acceleration**.

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