1. Write the position vector

The particle moves in a plane with constant $z$-coordinate: $z=7\text{ cm} = \text{constant}$

Given: $x=3t, \qquad y=5t^3$

So the position vector is $\vec r(t)=3t\,\hat{x}+5t^3\,\hat{y}+7\,\hat{z}$

2. Find velocity by differentiating position

$\vec v(t)=\frac{d\vec r}{dt}$

Differentiate each component: $\frac{dx}{dt}=3, \qquad \frac{dy}{dt}=15t^2, \qquad \frac{dz}{dt}=0$

Hence, $\vec v(t)=3\hat{x}+15t^2\hat{y}$

3. Find acceleration by differentiating velocity

$\vec a(t)=\frac{d\vec v}{dt}$

Differentiate again: $\frac{d}{dt}(3)=0, \qquad \frac{d}{dt}(15t^2)=30t$

Thus, $\vec a(t)=30t\hat{y}$

4. Evaluate at $t=1\text{ s}$

$\vec a(1)=30(1)\hat{y}=30\hat{y}$

5. Compare with options

• A: $-30\hat{y}$ ❌
• B: $30\hat{y}$ ✅
• C: $3\hat{x}+15\hat{y}$ ❌ (this is not acceleration)
• D: $3\hat{x}+15\hat{y}+7\hat{z}$ ❌ (this is not acceleration)

Therefore, the correct answer is Option B.