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. `-5` m/s
B. `-3` m/s
C. 6 m/s
D. 2 m/s

Answer

To find the initial velocity of the particle moving along the X-axis, we start with the given position function:

\[ x(t) = 2 - 5t + 6t^2 \]

### Step 1: Differentiate the position function
To find the velocity, we need to differentiate the position function with respect to time \( t \). The velocity \( v(t) \) is given by:

\[ v(t) = \frac{dx}{dt} \]

Differentiating \( x(t) \):

\[ \frac{dx}{dt} = \frac{d}{dt}(2) + \frac{d}{dt}(-5t) + \frac{d}{dt}(6t^2) \]

### Step 2: Apply the differentiation rules
Using the rules of differentiation:

- 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).

Putting it all together:

\[ v(t) = 0 - 5 + 12t \]

Thus, we have:

\[ v(t) = 12t - 5 \]

### Step 3: Calculate the initial velocity
The initial velocity is found by evaluating the velocity function at \( t = 0 \):

\[ v(0) = 12(0) - 5 \]

Calculating this gives:

\[ v(0) = 0 - 5 = -5 \, \text{m/s} \]

### Conclusion
The initial velocity of the particle is:

\[ \text{Initial Velocity} = -5 \, \text{m/s} \]

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