The x and y coordinates of the particle at any time are `x=5t-2t^(2) and y=10t` respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2s is:

Options

A. `5m//s^(2)`
B. `-4m//s^(2)`
C. `-8m//s^(2)`
D. `0`

Explanation

To find the acceleration of the particle at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Write down the equations for x and y coordinates
The coordinates of the particle are given as:
\[ x = 5t - 2t^2 \]
\[ y = 10t \]

### Step 2: Differentiate the x-coordinate to find velocity in the x-direction
To find the velocity in the x-direction (\( v_x \)), we differentiate \( x \) with respect to \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(5t - 2t^2) = 5 - 4t \]

### Step 3: Differentiate the y-coordinate to find velocity in the y-direction
To find the velocity in the y-direction (\( v_y \)), we differentiate \( y \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(10t) = 10 \]

### Step 4: Differentiate the velocity in the x-direction to find acceleration in the x-direction
To find the acceleration in the x-direction (\( a_x \)), we differentiate \( v_x \) with respect to \( t \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(5 - 4t) = -4 \]

### Step 5: Differentiate the velocity in the y-direction to find acceleration in the y-direction
To find the acceleration in the y-direction (\( a_y \)), we differentiate \( v_y \) with respect to \( t \):
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(10) = 0 \]

### Step 6: Calculate the resultant acceleration
The resultant acceleration (\( a \)) can be found using the Pythagorean theorem:
\[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle at \( t = 2 \) seconds is:
\[ \boxed{4 \, \text{m/s}^2} \]