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 is \(\vec{v} = \frac{d\vec{r}}{dt} = 4\hat{i} + 4t\hat{j}\). At \(t = 1\text{ s}\), \(\vec{v} = 4\hat{i} + 4\hat{j}\). Thus, magnitude \(v = \sqrt{4^2 + 4^2} = 4\sqrt{2}\text{ ms}^{-1}\) and angle \(\tan\theta = \frac{4}{4} = 1 ⇒ \theta = 45^\circ\) with respect to the x-axis.
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