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anonymous
 one year ago
the position of a particle as it moves along the axis is given for t>0 by x=(t^33t^2+6t)m,where t is in s.where is the particle when it achieves its minimum speed (after t=0)
anonymous
 one year ago
the position of a particle as it moves along the axis is given for t>0 by x=(t^33t^2+6t)m,where t is in s.where is the particle when it achieves its minimum speed (after t=0)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think u have to do double differentiation here ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@HS_CA @CallMeKiki @countrygirl1431 @carolinar7 @Camila1315 @Ivanskodje @isry98 @just_one_last_goodbye @kobi @Loser66 @lordhelix8th @LoneWolfKay

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SolomonZelman @ParthKohli

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Do you know about minima and maxima?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4if 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 tcoordinate 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 \(x(t_0)\)
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