A particle moves along X-axis in such a way that its coordinate X varies with time t according to the equation `x = (2-5t +6t^(2))m`. The initial velocity of the particle is

Options

A. `-5m//s`
B. `6m//s`
C. `-3m//s`
D. `3m//s`

Answer

To find the initial velocity of the particle, we need to follow these steps:

1. **Understand the equation of motion**: The position of the particle as a function of time is given by the equation:
\[ x(t) = 2 - 5t + 6t^2 \]

2. **Differentiate the position function**: The velocity of the particle is the derivative of the position function with respect to time. We can express this mathematically as:
\[ v(t) = \frac{dx}{dt} \]

3. **Calculate the derivative**: We will differentiate the position function \(x(t)\):
\[ v(t) = \frac{d}{dt}(2 - 5t + 6t^2) \]
- The derivative of a constant (2) is 0.
- The derivative of \(-5t\) is \(-5\).
- The derivative of \(6t^2\) is \(12t\) (using the power rule).

Therefore, we have:
\[ v(t) = 0 - 5 + 12t = -5 + 12t \]

4. **Find the initial velocity**: The initial velocity is the velocity at time \(t = 0\):
\[ v(0) = -5 + 12(0) = -5 \]

5. **Conclusion**: The initial velocity of the particle is:
\[ v(0) = -5 \, \text{m/s} \]

### Final Answer:
The initial velocity of the particle is \(-5 \, \text{m/s}\).

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