Question

If the displacement \( s \) (in metre) of a moving particle in terms of time \( t \) (in second) is \( s = t^3 - 6t^2 + 18t + 9 \), then the minimum velocity attained by the particle is

• A. \( 29~\text{ms}^{-1} \)
• B. \( 5~\text{ms}^{-1} \)
• C. \( 6~\text{ms}^{-1} \)
• D. \( 12~\text{ms}^{-1} \)

Hint:

To find minimum or maximum velocity, differentiate displacement to get velocity, then find critical points by setting derivative of velocity to zero. Confirm using second derivative.

Answer 1

The Correct Option is C

Step 1: Differentiate displacement to get velocity. \[ v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 18t + 9) = 3t^2 - 12t + 18 \] Step 2: Find critical points (minimum/maximum velocity). Set derivative of velocity to zero: \[ \frac{dv}{dt} = 6t - 12 = 0 \Rightarrow t = 2 \] Step 3: Find velocity at \( t = 2 \) \[ v = 3(2)^2 - 12(2) + 18 = 12 - 24 + 18 = 6~\text{ms}^{-1} \] Step 4: Confirm it's minimum by second derivative test. \[ \frac{d^2v}{dt^2} = 6>0 \Rightarrow \text{Minimum at } t = 2 \] % Final Answer \[ \boxed{6~\text{ms}^{-1}} \]