Question

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

• A. 16 m/s in y-direction
• B. 4 m/s in x-direction
• C. 9 m/s in z-direction
• D. 4 m/s in y-direction

Answer 1

Answer: (D)

4 m/s in y-direction

Answer 2

Step 1. Given Position Vector: $S = 2t^2 \hat{j} + 5t \hat{k}$

Step 2. Calculate Velocity Vector: The velocity vector $\vec{v}$ is the derivative of the position vector $S$ with respect to $t$:
$\vec{v} = \frac{dS}{dt} = \frac{d}{dt}(2t^2 \hat{j} + 5t \hat{k}) = (4t) \hat{j} + 5 \hat{k}$

Step 3. Substitute $t = 1$ to Find Velocity: At $t = 1$:
$\vec{v} = (4 \cdot 1) \hat{j} + 5 \hat{k} = 4 \hat{j} + 5 \hat{k}$

Step 4. Direction of Velocity: The y-component of velocity is 4 m/s, which matches option (4).
Thus, the magnitude and direction of velocity at $t = 1$ s is 4 m/s in the y-direction.