anonymous
  • anonymous
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)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@zepdrix @Miracrown
anonymous
  • anonymous
i think u have to do double differentiation here ...
anonymous
  • anonymous
@HS_CA @CallMeKiki @countrygirl1431 @carolinar7 @Camila1315 @Ivanskodje @isry98 @just_one_last_goodbye @kobi @Loser66 @lordhelix8th @LoneWolfKay

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anonymous
  • anonymous
@SolomonZelman @ParthKohli
anonymous
  • anonymous
@Nnesha @Abhisar
Abhisar
  • Abhisar
Do you know about minima and maxima?
anonymous
  • anonymous
no
Michele_Laino
  • 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 \(x(t_0)\)

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