Question

The displacement \(S\) of a particle measured from a fixed point \(O\) on a line is given by \[ S = t^3 - 16t^2 + 64t - 16. \] Then the time at which the displacement of the particle is maximum is

• A. 8
• B. 4
• C. \(\dfrac{8}{3}\)
• D. \(\dfrac{4}{3}\)

Hint:

To find the time of maximum displacement, take the derivative of displacement and apply the second derivative test.

Answer 1

The Correct Option is C

We are given:
\[ S = t^3 - 16t^2 + 64t - 16 \] Step 1: Find critical points
To find the time when displacement is maximum, take the derivative of \(S\) with respect to time \(t\):
\[ \frac{dS}{dt} = 3t^2 - 32t + 64 \] Set \(\frac{dS}{dt} = 0\) for maximum or minimum displacement:
\[ 3t^2 - 32t + 64 = 0 \] Step 2: Solve the quadratic
Use the quadratic formula:
\[ t = \frac{32 \pm \sqrt{(-32)^2 - 4(3)(64)}}{2 \cdot 3} = \frac{32 \pm \sqrt{1024 - 768}}{6} = \frac{32 \pm \sqrt{256}}{6} = \frac{32 \pm 16}{6} \Rightarrow t = \frac{48}{6} = 8, \quad \frac{16}{6} = \frac{8}{3} \] Step 3: Use second derivative test
Differentiate again:
\[ \frac{d^2S}{dt^2} = 6t - 32 \] - At \(t = 8\): \(\frac{d^2S}{dt^2} = 6(8) - 32 = 48 - 32 = 16 > 0\) → minimum - At \(t = \frac{8}{3}\): \(\frac{d^2S}{dt^2} = 6 \cdot \frac{8}{3} - 32 = 16 - 32 = -16 < 0\) → maximum Step 4: Final Answer
Maximum displacement occurs at \(t = \boxed{\dfrac{8}{3}}\)