Position of an ant ($\mathrm{S}$ in metres) moving in $\mathrm{Y}$-$\mathrm{Z}$ plane is given by $S=2 t^2 \hat{j}+5 \hat{k}$ (where $t$ is in second). The magnitude and direction of velocity of the ant at $\mathrm{t}=1 \mathrm{~s}$ will be :

<p>The position of an ant, denoted as $ \mathrm{S} $ in meters, moving in the $ \mathrm{Y} $-$ \mathrm{Z} $ plane is given by $ S = 2t^2 \hat{j} + 5 \hat{k} $, where $ t $ is in seconds. To determine the magnitude and direction of the ant's velocity at $ t = 1 $ second, we need to differentiate the position function with respect to time.</p> <p>The velocity $ \overrightarrow{\mathrm{v}} $ is given by:</p> <p>$$\overrightarrow{\mathrm{v}} = \frac{d\mathrm{S}}{dt} = \frac{d}{dt} (2t^2 \hat{j} + 5 \hat{k})$$</p> <p>On differentiating, we get:</p> <p>$\overrightarrow{\mathrm{v}} = 4t \hat{j}$</p> <p>At $ t = 1 $ second:</p> <p>$\overrightarrow{\mathrm{v}} = 4 \cdot 1 \hat{j} = 4 \hat{j}$</p> <p>Therefore, the magnitude of the velocity is $ 4 \mathrm{~m/s} $ and it is directed along the $ y $-axis.</p>