Find total displacement of the particle in motion

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SUMMARY

The total displacement of the particle described by the equation X = 3T^2 - T^3 over the first 4 seconds is -16 meters. The velocity function derived from the position equation is v = 6t - 3t^2, which equals zero at t = 2 seconds. The particle moves from 0 meters to -16 meters during this time, resulting in a total displacement of -16 meters. The discussion clarifies the distinction between total displacement and total distance, with the latter being 24 meters due to the particle's movement direction.

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hello everybody,
what is the total displacement of the particle at first 4 second?

equation (x) versus (t) : X = 3T^2 - T^3 where x is in meter and t in second.

my solution is : v= 6t-3t^2 , 6t-3t^2=0 , t=2 ,

X(2)= +4 , X(4)= -16 ,: displacement of the particle at first 4 second is: -16 m

BUT I'm not sure which my answer is right. please guide me.
 
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Yes, in the first 4 seconds, the particle moved from 0 to -16 and so the displacement is -16 m.

But the wording of the problem bothers me. What is total displacement?

Since the particle moved 4 meters to the right, then -20 meters to the left, the total distance (not displacement) the particle moved is 24 meters.
 

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