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 components
1. Differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(4t^2 + 5t + 16) = 8t + 5 \]
Thus, the x-component of velocity \( V_x = 8t + 5 \).

2. Differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(5t) = 5 \]
Thus, the y-component of velocity \( V_y = 5 \).

### Step 2: Write the velocity vector
The velocity vector \( \vec{V} \) can be expressed as:
\[ \vec{V} = V_x \hat{i} + V_y \hat{j} = (8t + 5) \hat{i} + 5 \hat{j} \]

### Step 3: Differentiate the velocity components to find acceleration components
1. Differentiate \( V_x \) with respect to \( t \):
\[ \frac{dV_x}{dt} = \frac{d}{dt}(8t + 5) = 8 \]
Thus, the x-component of acceleration \( a_x = 8 \).

2. Differentiate \( V_y \) with respect to \( t \):
\[ \frac{dV_y}{dt} = \frac{d}{dt}(5) = 0 \]
Thus, the y-component of acceleration \( a_y = 0 \).

### Step 4: Write the acceleration vector
The acceleration vector \( \vec{a} \) can be expressed as:
\[ \vec{a} = a_x \hat{i} + a_y \hat{j} = 8 \hat{i} + 0 \hat{j} \]

### Step 5: Calculate the magnitude of the acceleration
The magnitude of the acceleration \( a \) is given by:
\[ a = \sqrt{(a_x)^2 + (a_y)^2} = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8 \, \text{m/s}^2 \]

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

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