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 meter. Find the magnitude and direction of velocity \(v(t)\), at \(t = 1 \text{s}\), with respect to x-axis.
\(3\sqrt{2} \text{ms}^{-1}, 30^\circ\)
\(3\sqrt{2} \text{ms}^{-1}, 45^\circ\)
\(4\sqrt{2} \text{ms}^{-1}, 45^\circ\)
\(4\sqrt{2} \text{ms}^{-1}, 60^\circ\)
Solution:
Velocity \(\vec{v}(t) = \frac{d\vec{r}}{dt} = 4\hat{i} + 4\that{j}\). At \(t = 1 \text{s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2} \text{m/s}\). The angle with the x-axis is \(tan\theta = \frac{v_y}{v_x} = \frac{4}{4} = 1 ⇒
\theta = 45^\circ\).
Leave a Reply