# A particle

A particle moves in a straight line so that its velocity (m/s) at time t seconds is given by v(t) = 3t2-4t-4, t>0.

Initially the particle is 8 meters to the right of a fixed origin.

After how many seconds is the particle at the origin?

Result

t =  2 s

#### Solution:

$v(t)=3t^2-4t-4 \ \\ s(t)=\int v(t) dt \ \\ s(t)=t^3 - 2 \ t^2 -4t + C \ \\ C=8 \ \\ \ \\ s(t)=0 \ \\ t^3 - 2 \ t^2 -4t + 8=0 \ \\ t_{1}=-2 \ \\ t_{2}=2 \ \\ \ \\ t>0 \ \\ t=t_{2}=2 \ \text{s}$

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