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

The Correct Option is D

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.