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: Understanding the Concept:
Velocity is the rate of change of distance with respect to time. To find the velocity function, we need to differentiate the distance function \( s(t) \) with respect to time \( t \).
Step 2: Key Formula or Approach:
1. Velocity, \( v(t) = \frac{ds}{dt} \).
2. Set \( v(t) = 0 \) and solve for t.
Step 3: Detailed Explanation:
Based on the reconstruction, let the distance function be \( s(t) = t^2 - 12t + 8 \).
First, find the velocity function \( v(t) \) by differentiating \( s(t) \):
\[ v(t) = \frac{d}{dt}(t^2 - 12t + 8) \] \[ v(t) = 2t - 12 \] Next, we need to find the time 't' when the velocity becomes zero. Set \( v(t) = 0 \):
\[ 2t - 12 = 0 \] \[ 2t = 12 \] \[ t = \frac{12}{2} = 6 \] The velocity becomes zero when \( t = 6 \) seconds.
Step 4: Final Answer:
The velocity becomes zero when t is equal to 6.