Question

The displacement of a particle moving along x-axis is given as $(x=\frac{3}{4}t^{2}-12t+3)m$. The particle will come to rest in

• A. 4 s
• B. 5 s
• C. 16 s
• D. 8 s
• E. 6 s

Hint:

Rest $\Rightarrow v=0$; Acceleration = constant $\Rightarrow v = u + at$.

Answer 1

The Correct Option is D

Step 1: Concept
Velocity ($v$) is the first derivative of displacement ($x$) with respect to time ($t$).

Step 2: Meaning
"Coming to rest" means the velocity becomes zero ($v = 0$).

Step 3: Analysis
$v = \frac{dx}{dt} = \frac{d}{dt} (\frac{3}{4}t^{2}-12t+3) = \frac{3}{4}(2t) - 12 = 1.5t - 12$.
Set $v = 0$: $1.5t = 12 \Rightarrow t = \frac{12}{1.5} = 8$ s.

Step 4: Conclusion
Hence, the particle comes to rest at 8 seconds.
Final Answer: (D)