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 position equations
The position equations given are:
\[ x(t) = 2t + 4t^2 \]
\[ y(t) = 5t \]

### Step 2: Differentiate to find the velocity
To find the velocity, we differentiate the position equations with respect to time \( t \).

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

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

### Step 3: Differentiate to find the acceleration
Next, we differentiate the velocity equations to find the acceleration.

**For the x-direction:**
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(2 + 8t) = 8 \]

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

### Step 4: Calculate the net acceleration
The net acceleration \( a \) can be calculated using the Pythagorean theorem:
\[ a = \sqrt{a_x^2 + a_y^2} = \sqrt{8^2 + 0^2} = \sqrt{64} = 8 \, \text{m/s}^2 \]

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