If `s=t^3-4t^2+5` describes the motion of a particle, then its velocity when the acceleration vanishes, is
(a) `16/9` unit/sec
 (b) `-32/3` unit/sec
(c) `4/3` unit/sec             
(d)`-16/3` unit/sec

Explanation

To solve the problem, we will follow these steps:

1. **Identify the position function**: The position of the particle is given by the equation \( s = t^3 - 4t^2 + 5 \).

2. **Find the velocity function**: The velocity \( v \) of the particle is the first derivative of the position function with respect to time \( t \).
\[ v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 4t^2 + 5) \]
...