Particle 's position as a function of time is given by `x=-t^(2)+4t+4` find the maximum value of position coordinate of particle.

Answer

To find the maximum value of the position coordinate of the particle given by the function \( x(t) = -t^2 + 4t + 4 \), we can follow these steps:

### Step 1: Find the derivative of the position function
To determine the maximum value, we first need to find the derivative of the position function with respect to time \( t \).

\[ \frac{dx}{dt} = \frac{d}{dt}(-t^2 + 4t + 4) \]

Using the power rule for differentiation, we get:

\[ \frac{dx}{dt} = -2t + 4 \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:

\[ -2t + 4 = 0 \]

### Step 3: Solve for \( t \)
Now, we solve for \( t \):

\[ -2t = -4 \implies t = 2 \]

### Step 4: Find the maximum position by substituting \( t \) back into the original function
Now that we have the value of \( t \), we substitute it back into the original position function to find the maximum position:

\[ x(2) = - (2)^2 + 4(2) + 4 \]

Calculating this gives:

\[ x(2) = -4 + 8 + 4 = 8 \]

### Conclusion
Thus, the maximum value of the position coordinate of the particle is:

\[ \boxed{8} \]

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