Question

The distance travelled by a moving particle is given by $s=\frac{t^{2}}{2}-6t+8$, where $t$ denotes the time in seconds. The velocity becomes zero when $t$ is equal to:

• A. 1
• B. 4
• C. 3
• D. 6
• E. 8

Hint:

Velocity is zero at the turning points of the displacement-time graph.

Answer 1

The Correct Option is D

Step 1: Concept
Velocity $v$ is the derivative of distance $s$ with respect to time $t$ ($v = \frac{ds}{dt}$).

Step 2: Analysis
Differentiating $s = \frac{t^2}{2} - 6t + 8$: $v = \frac{2t}{2} - 6 = t - 6$.

Step 3: Calculation
Set velocity to zero: $t - 6 = 0 \implies t = 6$. Final Answer: (D)