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$$
\begin{array}{l}{\text { A rocket starts from rest and moves upward from the surface of the earth. For the first } 10.0 \text { s of its }} \\ {\text { motion, the vertical acceleration of the rocket is given by ay } a y=\left(2.8 \mathrm{m} / \mathrm{s}^{\wedge}3\right) \mathrm{t} \text { , where the } \mathrm{y} \text { -direction }} \\ {\text { is upward. (a) What is the height of the rocket above the surface of the earth at } t=10 \mathrm{s} ?(\mathrm{b}) \text { what }} \\ {\text { is the speed of the rocket when it is } 325 \mathrm{m} \text { above the surface of the earth? }}\end{array}
$$

$$
v_{0}=0 \quad \uparrow+\quad t=10s
$$

$$
v_{y}(t)= \int_{0}^{t} a_{y} d t \ , \quad y-y_{0}= \int_{0}^{t} v_{y} d t
$$

\(∴ v_{y}(t)=\int_{0}^{t} a y d t=\int_{0}^{t}(2.8) t d t=1.4 t^{2} \)

$$
y - y_{0}=\int_{0}^{t} v_{y}(t) d t=\int_{0}^{t} 1.4 t^{2} d t=0.4667 t^{3}\
$$

$$
@ \ t= 10s \longrightarrow y-y_{0}= 0.4667(10)^{3}=467 \mathrm{m}
$$

$$
y-y_{0}=325 m \rightarrow 325=0.4667 t^{3} \rightarrow t=8.864s
$$

$$
V_{y}=1.4 t^{2}=1.4(8.864)^{2}=110 \mathrm{m} / \mathrm{s}
$$

$$
\begin{array}{l}{\text { The acceleration of a motorcycle is given by } a x(t)=A t-B t^{\wedge} 2 \text { Where } A=1.5 \mathrm{m} / \mathrm{s}^{\wedge} 3 \text { and }} \\ {B=0.12 \mathrm{m} / \mathrm{s}^{\wedge} 4 . \text { The motorcycle is at rest at the origin at time } t=0 \text { (a) Find its position }} \\ {\text { and velocity as functions of time. (b) Calculate the maximum velocity it attains. }}\end{array}
$$

$$
V_{x}=V_{0x}+\int_{0}^{t} a_{x} d t
$$

$$
V_{x}=V_{0 x}+\int_{0}^{t}\left(A t-B t^{2}\right) d t=V_{0 x}+\frac{1}{2} A t^{2}-\frac{1}{3} B t^{3}
$$

$$
\text { At Rest } \rightarrow t=0, \quad V_{0x}=0
$$

$$
\rightarrow v_{x}=\frac{1}{2} A t^{2}-\frac{1}{3} B t^{3}=\frac{1}{2}(1.5) t^{2}-\frac{1}{3}(0.12) t^{3}
$$

\(∴ V_{x}=\left(0.75 \mathrm{m} / \mathrm{s}^{3}\right) \mathrm{t}^{2}-\left(0.04 \mathrm{m} / \mathrm{s}^{4}\right) \mathrm{t}^{3} \)

$$
x=x_{0}+\int_{0}^{t}\left(\frac{1}{2} A t^{2}-\frac{1}{3} B t^{3}\right) d t=x_{0}+\frac{1}{6} A t^{3}-\frac{1}{12} B t^{4}
$$

$$
\text { At Origin } \longrightarrow t=0, \quad  x_{0}=0
$$

$$
\rightarrow x=\frac{1}{6} A t^{3}-\frac{1}{12} B t^{4}=\frac{1}{6}(1.5) t^{3}-\frac{1}{12}(0.12) t^{4}
$$

\(∴ x=\left(0.25 \mathrm{m} / \mathrm{s}^{3}\right) \mathrm{t}^{3}-\left(0.01 \mathrm{m} / \mathrm{s}^{4}\right) \mathrm{t}^{4} \)

$$
a_{x} = \frac{d v_{x}}{d t}=0 \rightarrow A t-B t^{2}=0
$$

$$
\text { One root is } \mathrm{t}=0, \quad \text { At this Time } \quad \mathrm{V}_{x}=0 \text { and } \mathrm{NOT} \text { the Max. }
$$

$$
\text { At This Time } \rightarrow V_{x}=(0.75) t^{2}-(0.04) t^{3}
$$

$$
\rightarrow V_{x}=(0.75)(12.5)^{2}-(0.04)(12.5)^{3}=117.2-78.1=39.1 \mathrm{m} / \mathrm{s}
$$

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