If `x = 5t + 3t^(2)` and `y = 4t` are the x and y co-ordinates of a particle at any time t second where x and y are in metre, then the acceleration of the particle

Options

A. is zero throughout its motion
B. is a constant throughout its motion
C. depends only on its y component
D. varies along both x and y direction

Answer

To find the acceleration of the particle given the equations for its position in the x and y coordinates, we can follow these steps:

### Step 1: Write down the equations for x and y
The position of the particle is given by:
- \( x(t) = 5t + 3t^2 \)
- \( y(t) = 4t \)

### Step 2: Find the velocity components
The velocity components can be found by differentiating the position functions with respect to time \( t \).

- For the x-component of velocity \( v_x \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(5t + 3t^2) = 5 + 6t \]

- For the y-component of velocity \( v_y \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(4t) = 4 \]

### Step 3: Write the velocity vector
The velocity vector \( \vec{v} \) can be expressed as:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = (5 + 6t) \hat{i} + 4 \hat{j} \]

### Step 4: Find the acceleration components
The acceleration components can be found by differentiating the velocity functions with respect to time \( t \).

- For the x-component of acceleration \( a_x \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(5 + 6t) = 6 \]

- For the y-component of acceleration \( a_y \):
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(4) = 0 \]

### Step 5: Write the acceleration vector
The acceleration vector \( \vec{a} \) can be expressed as:
\[ \vec{a} = a_x \hat{i} + a_y \hat{j} = 6 \hat{i} + 0 \hat{j} \]

### Step 6: Conclusion about the acceleration
The acceleration of the particle is:
\[ \vec{a} = 6 \hat{i} \]
This indicates that the acceleration is constant and directed along the x-axis.

### Final Answer
The acceleration of the particle is constant and equal to \( 6 \, \text{m/s}^2 \) in the x-direction.

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