$($II$)$ A particle moving in one dimension experiences the piecewise acceleration, $a(t),$ shown in the adjacent figure. Time is measured in seconds and acceleration in $\frac{m}{s^2}.$ The open circles indicate instants at which the acceleration changed abruptly $-$ effectively instantaneously.

At $t=0s$ the particle is at $x_0=-2\mathrm{.}5m$ and is moving with velocity $v_0=-1\frac{m}{s}.$

Our goal, in this question, is to determine the particle's position and velocity at $t=10s.($R$1)($i$)$ Determine an expression for the particle's velocity $[$expressed as a function of time$]$ during the time interval tin$[0,5]s.$

$\frac{?}{\text{b}\text{a}\text{r}}(v)=\frac{s}{t}$

$($ii$)$ Compute the velocity of the particle at the instant $t=5s.,$vec$(v)(t)=$vec$(v{)}_0+$vec$(a).$

$($iii$)$ Determine an expression for the particle's position during the time interva tin$[0,5]s.$

$($iv$)$ Compute the position of the particle at the instant $t=5s.$

$P=\frac{-0\mathrm{.}2}{s}=0\mathrm{.}04(\frac{(m)}{(s)})-2\mathrm{.}5$:$0\mathrm{.}04=0\mathrm{.}1m(\frac{1}{3})$

$($R$2)$ Let tilde$(t)=t-5$ denote the time that the particle spends in the second acceleration regime.

$($i$)$ Determine an expression for the particle's velocity during the time interval tilde$(t)in[0,5]s.$

$\frac{?}{\text{b}\text{a}\text{r}}(v)=\frac{s}{t}=,\frac{-0\mathrm{.}2}{s}=-0\mathrm{.}04\frac{m}{s}$