## anonymous one year ago 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)

• This Question is Open
1. anonymous

@zepdrix @Miracrown

2. anonymous

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

3. anonymous

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

4. anonymous

@SolomonZelman @ParthKohli

5. anonymous

@Nnesha @Abhisar

6. Abhisar

Do you know about minima and maxima?

7. anonymous

no

8. Michele_Laino

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 $$x(t_0)$$