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If s = t3 – 4t2 + 5 describes the motion of a particle, then its velocity when the acceleration vanishes, is
If we are given the distance travelled(s) as a function of time, we can calculate velocity(v) and acceleration(a) by
and a![]()
s(t)= t3 – 4t2 + 5
differentiating w.r.t. time, we get
![]()
Differentiating again w.r.t. time, we get
a![]()
Given that a=0 ⇒ 6t-8=0 or
units
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