The x and y coordinates of a particle at any time t are given by `x = 2t + 4t^2 and y = 5t` , where x and y are in metre and t in second. The acceleration of the particle at t = 5 s is

Options

A. `40 ms ^(-2)`
B. `20 ms ^(-2)`
C. `8 ms ^(-2)`
D. zero

Answer

To find the acceleration of the particle at \( t = 5 \) seconds, we need to follow these steps:

### Step 1: Find the velocity in the x-direction
The x-coordinate of the particle is given by:
\[ x = 2t + 4t^2 \]
To find the velocity in the x-direction (\( v_x \)), we differentiate \( x \) with respect to \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(2t + 4t^2) \]
Using the power rule of differentiation:
\[ v_x = 2 + 8t \]

### Step 2: Find the acceleration in the x-direction
Next, we find the acceleration in the x-direction (\( a_x \)) by differentiating \( v_x \) with respect to \( t \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(2 + 8t) \]
Differentiating gives:
\[ a_x = 8 \]

### Step 3: Find the velocity in the y-direction
The y-coordinate of the particle is given by:
\[ y = 5t \]
To find the velocity in the y-direction (\( v_y \)), we differentiate \( y \) with respect to \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(5t) \]
This gives:
\[ v_y = 5 \]

### Step 4: Find the acceleration in the y-direction
Now, we find the acceleration in the y-direction (\( a_y \)) by differentiating \( v_y \) with respect to \( t \):
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(5) \]
Since the derivative of a constant is zero:
\[ a_y = 0 \]

### Step 5: Combine the results
At \( t = 5 \) seconds, the acceleration of the particle is:
\[ a_x = 8 \, \text{m/s}^2 \quad \text{and} \quad a_y = 0 \, \text{m/s}^2 \]
Thus, the total acceleration of the particle is:
\[ \text{Acceleration} = (a_x, a_y) = (8, 0) \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle at \( t = 5 \) seconds is:
\[ \boxed{8 \, \text{m/s}^2} \]