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 ms^(-2)`
C. `20 ms^(-2)`
D. `40 ms^(-2)`

Explanation

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

### Step 1: Identify the position functions
The position of the particle is given by:
- \( x(t) = 7t + 4t^2 \)
- \( y(t) = 5t \)

### Step 2: Differentiate the position functions to find velocity
We need to find the first derivatives of \( x(t) \) and \( y(t) \) with respect to time \( t \) to get the velocity components:
- \( v_x = \frac{dx}{dt} = \frac{d}{dt}(7t + 4t^2) = 7 + 8t \)
- \( v_y = \frac{dy}{dt} = \frac{d}{dt}(5t) = 5 \)

### Step 3: Differentiate the velocity functions to find acceleration
Now, we differentiate the velocity functions to find the acceleration components:
- \( a_x = \frac{d^2x}{dt^2} = \frac{d}{dt}(7 + 8t) = 8 \)
- \( a_y = \frac{d^2y}{dt^2} = \frac{d}{dt}(5) = 0 \)

### Step 4: Calculate the magnitude of acceleration
The magnitude of the total acceleration \( a \) can be found using the formula:
\[ a = \sqrt{a_x^2 + a_y^2} \]
Substituting the values we found:
\[ a = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8 \, \text{m/s}^2 \]

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

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