For a particle moving along x(t) = t³ − 3t² + 5, the velocity at t = 2…

Question

For a particle moving along x(t) = t³ − 3t² + 5, the velocity at t = 2 is:

Accepted Answer

D. $0$ m/s (3(4) − 6(2) = 0)

We have v(t) = dx/dt.
So v(t) = 3t² − 6t.
At t = 2: v = 3(4) − 6(2) = 12 − 12 = 0. So at t = 2, the particle is instantaneously at rest.
Could be turning point in motion.

Answer

$0$ (incorrectly assumed at rest)

Answer

$3$ m/s (omitted negative term)

Answer

$2$ m/s (wrong calculation)

For a particle moving along x(t) = t³ − 3t² + 5, the velocity at t = 2 is:

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