Question

Position of a particle is given by $\vec{r} = v_y\hat{i} + t^4\hat{j}$ (where t is time in second). Find magnitude of acceleration of the particle at t = 2 seconds. ($v_y$ represents velocity along y-axis)

• A. 24$\sqrt{2}$ m/s²
• B. 48$\sqrt{2}$ m/s²
• C. 12$\sqrt{3}$ m/s²
• D. 16$\sqrt{2}$ m/s²

Answer 1

Answer: (B)

48$\sqrt{2}$ m/s²

Answer 2

The position vector of the particle is given by $\vec{r} = v_y\hat{i} + t^4\hat{j}$. This implies that the components of the position vector are $x(t) = v_y$ and $y(t) = t^4$. The problem states that "$v_y$ represents velocity along y-axis", meaning $v_y = \frac{dy}{dt}$. Given $y(t) = t^4$, we find $v_y = \frac{dy}{dt} = 4t^3$. Therefore, the x-component of the position vector is $x(t) = 4t^3$. The position vector as a function of time is $\vec{r}(t) = 4t^3\hat{i} + t^4\hat{j}$. The velocity vector is obtained by differentiating the position vector with respect to time: $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(4t^3\hat{i} + t^4\hat{j}) = 12t^2\hat{i} + 4t^3\hat{j}$. The acceleration vector is obtained by differentiating the velocity vector with respect to time: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(12t^2\hat{i} + 4t^3\hat{j}) = 24t\hat{i} + 12t^2\hat{j}$. At $t = 2$ seconds, the acceleration vector is: $\vec{a}(2) = 24(2)\hat{i} + 12(2^2)\hat{j} = 48\hat{i} + 48\hat{j}$ m/s². The magnitude of the acceleration is $|\vec{a}(2)| = \sqrt{(48)^2 + (48)^2} = \sqrt{2 \times 48^2} = 48\sqrt{2}$ m/s².