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

Answer

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

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

### Step 2: Differentiate the x-coordinate to find velocity in the x-direction
To find the velocity in the x-direction, we differentiate \( x(t) \) with respect to \( t \):
\[ v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(7t + 4t^2) = 7 + 8t \]

### Step 3: Differentiate the y-coordinate to find velocity in the y-direction
Next, we differentiate \( y(t) \) with respect to \( t \):
\[ v_y(t) = \frac{dy}{dt} = \frac{d}{dt}(5t) = 5 \]

### Step 4: Differentiate the velocity in the x-direction to find acceleration in the x-direction
Now, we differentiate \( v_x(t) \) to find the acceleration in the x-direction:
\[ a_x(t) = \frac{dv_x}{dt} = \frac{d}{dt}(7 + 8t) = 8 \]

### Step 5: Differentiate the velocity in the y-direction to find acceleration in the y-direction
Similarly, we differentiate \( v_y(t) \) to find the acceleration in the y-direction:
\[ a_y(t) = \frac{dv_y}{dt} = \frac{d}{dt}(5) = 0 \]

### Step 6: Combine the accelerations to find the net acceleration
The net acceleration of the particle is given by the vector sum of the accelerations in the x and y directions:
\[ \text{Net Acceleration} = \sqrt{a_x^2 + a_y^2} = \sqrt{(8)^2 + (0)^2} = 8 \, \text{m/s}^2 \]

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