The x and y coordinates of a particle at any time t are given by `x=7t+4t^2` and `y=5t`, where x and t is seconds. The acceleration of particle at `t=5`s is

Options

A. zero
B. 8 `(m)/(s^2)`
C. 20 `(m)/(s^2)`
D. 40 `(m)/(s^2)`

Explanation

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

### Step 1: Identify the equations of motion
The x and y coordinates of the particle as functions of time \( t \) are given by:
\[ x(t) = 7t + 4t^2 \]
\[ y(t) = 5t \]

### Step 2: Find the velocity components
To find the acceleration, we first need to find the velocity components in the x and y directions. The velocity is the first derivative of the position with respect to time.

- For the x-component of velocity \( v_x \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(7t + 4t^2) = 7 + 8t \]

- For the y-component of velocity \( v_y \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(5t) = 5 \]

### Step 3: Find the acceleration components
The acceleration is the derivative of velocity with respect to time, which is the second derivative of position.

- For the x-component of acceleration \( a_x \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(7 + 8t) = 8 \]

- For the y-component of acceleration \( a_y \):
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(5) = 0 \]

### Step 4: Evaluate the acceleration at \( t = 5 \) seconds
From the calculations above, we find:
\[ a_x = 8 \, \text{m/s}^2 \]
\[ a_y = 0 \, \text{m/s}^2 \]

### Step 5: Calculate the resultant acceleration
The resultant acceleration \( a \) can be calculated using the Pythagorean theorem:
\[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8 \, \text{m/s}^2 \]

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