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$$
\begin{array}{l}{\text { 12-9. The acceleration of a particle as it moves along a }} \\ {\text { straight line is given by } a=(2 t-1) \mathrm{m} / \mathrm{s}^{2} \text { , where } t \text { is in }} \\ {\text { seconds. If } s=1 \mathrm{m} \text { and } v=2 \mathrm{m} / \mathrm{s} \text { when } t=0 \text { , determine }} \\ {\text { the particle's velocity and position when } t=6 \mathrm{s} \text { . Also, }} \\ {\text { determine the total distance the particle travels during this }} \\ {\text { time period. }}\end{array}
$$

$$
a=(2 t-1) m / s^{2}
$$

$$
\left.\begin{array}{l}{s=1m} \\ {v=2 m / s}\end{array}\right\} t=0
$$

$$
t= 6 \mathrm{s}
$$

$$
q=\frac{d v}{d t} \Rightarrow d v=a d t
$$

$$
\int_{2}^{v} d v=\int_{0}^{t}(2 t-1) d t \Rightarrow v-2=\frac{2t^2}{2}-t \Rightarrow v=t^2-t_{t^2} \rightarrow (1)
$$

$$
v=\frac{d s}{d t} \Rightarrow d s= v dt \Rightarrow \int_1^sds=\int_0^t(t^2-t_{t^2})dt \Rightarrow s=\frac{1}{3}t^3-\frac{1}{2}t^2+2t+1 \rightarrow (2)
$$

$$
v = 32 \mathrm{m} / \mathrm{s}, \  \mathrm{S}=67 \mathrm{m}
$$

$$
d=67-1
$$

$$
=66 m
$$

$$
\begin{array}{l}{\text { 12-22. A sandbag is droppod from a balloon which is }} \\ {\text { ascending vertically at a constant speed of } 6 \text { m/ s. If the bag }} \\ {\text { is released with the same upward velocity of } 6 \text { m/ s when }} \\ {\text { t=0 and hits the ground when } t=8 \text {s, determine the speed }} \\ {\text { of the bag as it hits the ground and the altitude of the }} \\ {\text { balloon at this instant. }}\end{array}
$$

\( \left.\begin{array}{l}{v=6 m / s \Rightarrow {t=0}} \\ {t=8s }\end{array}\right\} \begin{array}{l}{v=v_{0}+a_{c}t} \\ {(+ \downarrow) g=9.81 \mathrm{m/s}^{2}}\end{array} \)

$$
(+ \downarrow) v=v_{0}+g t
$$

\( =(-6)+(9.81)*8 ≃ 72.5 \mathrm{m} / \mathrm{s}  \)

$$
 S=s_0+v_0 t+\frac{1}{2} a_{c} t^{2} \Rightarrow h=h_{0} +V_0 t+\frac{1}{2} g t^{2}
$$

\(∴ h=0+(-6)(8)+\frac{1}{2}(9.81)(8)^{2}\)

\(≃265.92 \mathrm{m} \)

\( V=\frac{h^{\prime}}{t} \Rightarrow ∴ h^{\prime}=v \cdot t =6 * 8=48 \mathrm{m} \)

$$
h t=h+h^{\prime}=265.92+48 \simeq 314 m
$$

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