Question 37: Motion in 2D & Vectors

Question

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 :

Accepted Answer

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.
The velocity $ \overrightarrow{\mathrm{v}} $ is given by:
$$\overrightarrow{\mathrm{v}} = \frac{d\mathrm{S}}{dt} = \frac{d}{dt} (2t^2 \hat{j} + 5 \hat{k})$$
On differentiating, we get:
$$\overrightarrow{\mathrm{v}} = 4t \hat{j}$$
At $ t = 1 $ second:
$$\overrightarrow{\mathrm{v}} = 4 \cdot 1 \hat{j} = 4 \hat{j}$$
Therefore, the magnitude of the velocity is $ 4 \mathrm{~m/s} $ and it is directed along the $ y $-axis.

Answer

$$16 \mathrm{~m} / \mathrm{s}$$ in $$y$$-direction

Answer

$$4 \mathrm{~m} / \mathrm{s}$$ in $$x$$-direction

Answer

$$9 \mathrm{~m} / \mathrm{s}$$ in $$\mathrm{z}$$-direction

Answer

$$4 \mathrm{~m} / \mathrm{s}$$ in $$y$$-direction

Motion in 2D & Vectors

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 :

Choose the correct answer: