A particle moves in an XY-plane in such a way that its x and y-coordinates vary with time according to
`x(t)=t^(3)-32t, y(t)=5t^(2)+12`
Find the acceleration of the particle, if t = 3 s.

Options

A. `9 hat(i)+5 hat(j)`
B. `18hat(i)+10hat(j)`
C. `18hat(i)-5hat(j)`
D. `-18hat(i)+10hat(j)`

Explanation

To find the acceleration of the particle at \( t = 3 \) seconds, we will follow these steps:

### Step 1: Differentiate the position functions to find velocity
The position functions are given as:
\[ x(t) = t^3 - 32t \]
\[ y(t) = 5t^2 + 12 \]

We need to find the velocity in the x-direction \( v_x(t) \) and y-direction \( v_y(t) \).

**For \( v_x(t) \)**:
\[ v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 32t) \]
Differentiating:
\[ v_x(t) = 3t^2 - 32 \]

**For \( v_y(t) \)**:
\[ v_y(t) = \frac{dy}{dt} = \frac{d}{dt}(5t^2 + 12) \]
Differentiating:
\[ v_y(t) = 10t \]

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

**For \( a_x(t) \)**:
\[ a_x(t) = \frac{dv_x}{dt} = \frac{d}{dt}(3t^2 - 32) \]
Differentiating:
\[ a_x(t) = 6t \]

**For \( a_y(t) \)**:
\[ a_y(t) = \frac{dv_y}{dt} = \frac{d}{dt}(10t) \]
Differentiating:
\[ a_y(t) = 10 \]

### Step 3: Evaluate acceleration at \( t = 3 \) seconds
Now, we will substitute \( t = 3 \) seconds into the acceleration functions.

**For \( a_x(3) \)**:
\[ a_x(3) = 6 \times 3 = 18 \]

**For \( a_y(3) \)**:
\[ a_y(3) = 10 \]

### Step 4: Write the acceleration vector
The acceleration vector \( \vec{a} \) can be expressed as:
\[ \vec{a} = a_x \hat{i} + a_y \hat{j} \]
Substituting the values we found:
\[ \vec{a} = 18 \hat{i} + 10 \hat{j} \]

### Final Answer
The acceleration of the particle at \( t = 3 \) seconds is:
\[ \vec{a} = 18 \hat{i} + 10 \hat{j} \]

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