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

Options

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

Explanation

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

### Step 1: Find the expressions for velocity
The velocity in the x-direction (\( v_x \)) and y-direction (\( v_y \)) can be found by differentiating the position functions with respect to time \( t \).

1. **Differentiate \( x(t) \)**:
\[ x(t) = 7t + 4t^2 \]
\[ v_x = \frac{dx}{dt} = 7 + 8t \]

2. **Differentiate \( y(t) \)**:
\[ y(t) = 5t \]
\[ v_y = \frac{dy}{dt} = 5 \]

### Step 2: Find the expressions for acceleration
The acceleration in the x-direction (\( a_x \)) and y-direction (\( a_y \)) can be found by differentiating the velocity functions with respect to time \( t \).

1. **Differentiate \( v_x(t) \)**:
\[ v_x = 7 + 8t \]
\[ a_x = \frac{dv_x}{dt} = 8 \]

2. **Differentiate \( v_y(t) \)**:
\[ v_y = 5 \]
\[ a_y = \frac{dv_y}{dt} = 0 \]

### Step 3: Calculate the total acceleration
The total acceleration vector \( \mathbf{a} \) can be expressed as:
\[ \mathbf{a} = (a_x, a_y) = (8, 0) \]

### Step 4: Find the magnitude of acceleration
The magnitude of the 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 \( t = 5 \, \text{s} \) is \( 8 \, \text{m/s}^2 \).

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