The position of a particle is given by $\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}$, where $t$ is in seco

Question

The position of a particle is given by $\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}$, where $t$ is in seconds and $r$ in metres. Find the magnitude and direction of the velocity $v(t)$, at $t=1$ s, with respect to the x-axis.

Accepted Answer

1.  **Find Velocity Vector:**
    Velocity $\vec{v}(t)$ is the time derivative of the position vector $\vec{r}(t)$.
    Given $\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}$.
    $$ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(4t)\hat{i} + \frac{d}{dt}(2t^2)\hat{j} + \frac{d}{dt}(5)\hat{k} $$
    $$ \vec{v}(t) = 4\hat{i} + 4t\hat{j} $$

2.  **Calculate Velocity at t = 1 s:**
    Substitute $t = 1$ into the velocity equation:
    $$ \vec{v}(1) = 4\hat{i} + 4(1)\hat{j} = 4\hat{i} + 4\hat{j} 	ext{ ms}^{-1} $$

3.  **Calculate Magnitude:**
    The magnitude of the velocity vector is:
    $$ |\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 4^2} $$
    $$ |\vec{v}| = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} 	ext{ ms}^{-1} $$

4.  **Calculate Direction:**
    The angle $	heta$ with the x-axis is given by:
    $$ 	an 	heta = \frac{v_y}{v_x} = \frac{4}{4} = 1 $$
    $$ 	heta = 	an^{-1}(1) = 45^{\circ} $$
    Thus, the magnitude is $4\sqrt{2} 	ext{ ms}^{-1}$ and the direction is $45^{\circ}$ with the x-axis.

Step-by-Step Solution

Practice All PHYSICS Questions