Question

If displacement of particle moving in straight line is \[ S=t^3-3t^2+3t-4 \] then time interval in which \(S\) is increasing is

• A. only \((1,\infty)\)
• B. only \([0,1)\)
• C. \([0,\infty)\) only
• D. \([3,\infty)\)

Hint:

For motion problems, displacement increases exactly when velocity \(v=\frac{dS}{dt}\) becomes positive.

Answer 1

The Correct Option is A

Concept: A function increases when its derivative is positive. Thus check \[ \frac{dS}{dt}>0 \]

Step 1: Differentiate displacement.
\[ \frac{dS}{dt} = 3t^2-6t+3 \] \[ = 3(t^2-2t+1) \] \[ = 3(t-1)^2 \]

Step 2: Analyze sign.
Since square is always nonnegative, \[ (t-1)^2\ge0 \] Thus \[ \frac{dS}{dt}\ge0 \] At \[ t=1 \] derivative zero. Positive elsewhere. Hence increasing for \[ (1,\infty) \] Thus \[ \boxed{(1,\infty)} \]