Question

The position of a particle moving along X axis given by X = (-2t³ + 3t² + 5)m . The acceleration of the particle at the instant its velocity becomes zero is

Answer 1

Explanation:

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Answer 2

Answer:

Explanation:

$x= (- 2t^3+3t^2+5)m$

$v = \frac{dx}{dt} = -6t^2 +6t$

$0 = -6t^2-6t$

$0 = t(-6t+6)$

either t be 0 or 1.

$a = \frac{dv}{dt} =-12t+6$

$if t=0; a =-12(0)+6=6$

$if t=1; a = -12(1)+6=-6$

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