the position of a particle as it moves along the axis is given for t0 by x=(t^3-3t^2+6t)m,where t is in s.where is the particle when it achieves its minimum speed (after t=0)

Question

the position of a particle as it moves along the axis is given for t>0 by x=(t^3-3t^2+6t)m,where t is in s.where is the particle when it achieves its minimum speed (after t=0)

anonymous · Answer

@zepdrix @Miracrown

anonymous · Answer

i think u have to do double differentiation here ...

anonymous · Answer

@HS_CA @CallMeKiki @countrygirl1431 @carolinar7 @Camila1315 @Ivanskodje @isry98  @just_one_last_goodbye @kobi @Loser66 @lordhelix8th @LoneWolfKay

anonymous · Answer

@SolomonZelman @ParthKohli

anonymous · Answer

@Nnesha @Abhisar

abhisar · Answer

Do you know about minima and maxima?

anonymous · Answer

no

michele_laino · Answer

if we compute the first derivative of your function $x(t)$, we get:
$$v(t)=\frac{{dx}}{{dt}} = 3{t^2} - 6t + 6$$
which is the equation of a parabola. Now the minimum speed occurs at the t-coordinate of the vertex of such parabola. Such value of $t$ is:
$$t_0 =  - \frac{{ - 6}}{{2 \cdot 3}} = ...?$$
now, please replace $t=1$ into formula of $x(t)$, or in other words, compute the quantity $x(t_0)$