Question

A particle is moving in a straight line such that its distance at any time \(t\) is given by \[ s=\frac{t^4}{4}-2t^3+4t^2+7, \] then its acceleration is minimum at \[ t=\_ \]

• A. 1
• B. 2
• C. 1/2
• D. 3/2

Hint:

Acceleration is minimum when its rate of change (jerk) is zero.

Answer 1

The Correct Option is B

Step 1: Concept
Acceleration $a$ is the second derivative of displacement $s$ with respect to time $t$. To find where $a$ is minimum, set $\frac{da}{dt} = 0$.

Step 2: Meaning
$v = \frac{ds}{dt} = t^{3} - 6t^{2} + 8t$. Then $a = \frac{dv}{dt} = 3t^{2} - 12t + 8$.

Step 3: Analysis
To minimize $a$, find $\frac{da}{dt} = 6t - 12$. Set $\frac{da}{dt} = 0 \implies 6t = 12 \implies t = 2$.

Step 4: Conclusion
The second derivative of acceleration is $\frac{d^{2}a}{dt^{2}} = 6$, which is positive, confirming a minimum at $t = 2$. Final Answer: (B)