x-coordinate of a particle moving along this axis is `x = (2+t^2 + 2t^3).` Here, x is in meres and t in seconds. Find (a) position of particle from where it started its journey, (b) initial velocity of particle and (c) acceleration of particle at `t=2s.`

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Given:
The x-coordinate of the particle is given by the equation:
\[ x(t) = 2 + t^2 + 2t^3 \]
where \( x \) is in meters and \( t \) is in seconds.

### (a) Position of the particle from where it started its journey

To find the initial position of the particle, we need to evaluate \( x(t) \) at \( t = 0 \):
\[ x(0) = 2 + (0)^2 + 2(0)^3 = 2 + 0 + 0 = 2 \text{ meters} \]

### (b) Initial velocity of the particle

The initial velocity can be found by differentiating the position function \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2 + t^2 + 2t^3) \]
Calculating the derivative:
\[ v(t) = 0 + 2t + 6t^2 = 2t + 6t^2 \]
Now, we find the initial velocity by evaluating \( v(t) \) at \( t = 0 \):
\[ v(0) = 2(0) + 6(0)^2 = 0 \text{ meters/second} \]

### (c) Acceleration of the particle at \( t = 2 \, \text{s} \)

To find the acceleration, we differentiate the velocity function \( v(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(2t + 6t^2) \]
Calculating the derivative:
\[ a(t) = 2 + 12t \]
Now, we find the acceleration at \( t = 2 \, \text{s} \):
\[ a(2) = 2 + 12(2) = 2 + 24 = 26 \text{ meters/second}^2 \]

### Summary of Results:
(a) The position of the particle from where it started its journey is **2 meters**.
(b) The initial velocity of the particle is **0 meters/second**.
(c) The acceleration of the particle at \( t = 2 \, \text{s} \) is **26 meters/second²**.