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A particle moving along a coordinate line at time $t=0$ is at position $3$ cm from the origin and travelling with a velocity of $7$cm/s. If the acceleration of the particle is given by $$a(t)=2-2(t+1)^{-3}.$$ Find the velocity and position of the particle as function of t.

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  • $\begingroup$ Integrate acceleration to get velocity. $\endgroup$ – coffeemath Jan 15 '18 at 12:18
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The integral of the acceleration with respect to $t$, the velocity is

$$v (t)=2t+(t+1)^{-2}+C_v.$$

For $t=0$, the velocity is $7$. So, $C_v=6$.

The integral of the velocity with respect to $t$, the position is

$$t^2-(t+1)^{-1}+6t+C_p.$$

For $t=0$, the position is $3$. So, $C_p=4$, that is,

$$s (t)=t^2-(t+1)^{-1}+6t+4.$$

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