Question 2

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

The position is given by $$\vec{S} = 2t^2 \hat{j} + 5\hat{k}$$.

Velocity is the time derivative of position:

$$\vec{v} = \frac{d\vec{S}}{dt} = 4t\hat{j} + 0\hat{k} = 4t\hat{j}$$

At $$t = 1$$ s:

$$\vec{v} = 4(1)\hat{j} = 4\hat{j}$$ m/s

The magnitude is $$4$$ m/s and the direction is along the y-direction.

The answer is $$4 \text{ m s}^{-1}$$ in y-direction, which corresponds to Option (4).

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