A particle moves in x-y plane according to the equations `x= 4t^2+ 5t+ 16 and y=5t` where x, y are in metre and t is in second. The acceleration of the particle is

Options

A. `8 m s^(-2)`
B. `12 m s^(-2)`
C. `14 m s^(-2)`
D. `16 m s^(-2)`

Answer

To find the acceleration of the particle moving in the x-y plane according to the equations \( x = 4t^2 + 5t + 16 \) and \( y = 5t \), we will follow these steps:

### Step 1: Differentiate the position equations to find velocity
The first step is to find the velocity components in the x and y directions by differentiating the position equations with respect to time \( t \).

1. Differentiate \( x \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(4t^2 + 5t + 16) = 8t + 5 \]

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

### Step 2: Differentiate the velocity equations to find acceleration
Next, we differentiate the velocity components to find the acceleration components.

1. Differentiate \( v_x \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(8t + 5) = 8 \]

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

### Step 3: Write the acceleration vector
The acceleration of the particle can be expressed as a vector in the x-y plane:
\[ \vec{a} = a_x \hat{i} + a_y \hat{j} = 8 \hat{i} + 0 \hat{j} \]

### Step 4: Calculate the magnitude of the acceleration
To find the magnitude of the acceleration vector:
\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2} = \sqrt{8^2 + 0^2} = \sqrt{64} = 8 \, \text{m/s}^2 \]

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

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