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$$
\begin{array}{l}{\text { A jet plane is flying at constant altitude. At time t1 }=0 \text { it has components of velocity } \mathrm{V}{x=90} \mathrm{m} / \mathrm{s}, \mathrm{Vy}=110 \mathrm{m} / \mathrm{s} \text { At time the }} \\ {\text { the components are } \mathrm{v} x=-170 \mathrm{m} / \mathrm{s} \mathrm{s} \text { thang }} \\ {\text { (a) Sketch the velocity vectors at } 1 \mathrm{t} \text { and t2, How do these two vectors differ } \mathrm{fime} \text { interval calculate }} \\ {\text { (b) the components of the average accelerating, and }} \\ {\text { (c) the magnitude and direction of the average acceleration }}\end{array}
$$

$$
a_{a v}=\frac{\Delta v}{\Delta t}
$$

$$
\left(a_{a v{g}}\right)_{x}=\frac{\Delta v_{x}}{\Delta t}
$$

$$
\left(a_{\text {avg }}\right)_ x=\frac{\left(V_{x}\right)_{2}-(V_ x)_{1}}{\Delta t}
$$

$$
=\frac{-170-90}{30-0}
$$

$$
=-8.67 \mathrm{m} / \mathrm{s}^{2}
$$

$$
\left(a_{a v{g}}\right)_ y=\frac{\Delta v_{y}}{\Delta t}
$$

$$
(a _{avg} )_ y=\frac{\left(v_{y}\right)_{2}-\left(v_{y}\right)_{1}}{\Delta t}
$$

$$
=\frac{40-110}{30-0}
$$

$$
=-2.33 \mathrm{m} / \mathrm{s}^{2}
$$

$$
\left[\left(a_{\left.a v{g}\right)_{x}}=-8.67 m\left/s^{2},\left(a_{a v{g}}\right)_ y=-2.33 m\right/ s^{2}\right]\right.
$$

$$
a=\sqrt{\left(a_{avg }\right)_{x}^{2}+\left(a_{avg}\right)_{y}^{2}}=\sqrt{(-8.67)^{2}+(-2.33)^{2}}
$$

$$
=8.98 \mathrm{m} / \mathrm{s}^{2}
$$

$$
\tan \alpha=\frac{\left(a_{av}\right)_ y}{\left(a_{av}\right)_{x}}=\frac{-2.33}{-8.67}=0.269
$$

$$
\alpha=\tan ^{-1}(0.269)=15^{\circ}
$$

$$
\rightarrow \alpha=180^{\circ}+15^{\circ}=195^{\circ}
$$

$$
\begin{array}{l}{\text { A dog running in an open field has components of velocity } \mathrm{V}_{\mathrm{X}}=2.6 \mathrm{m} / \mathrm{s} \text { And } \mathrm{V}_{\mathrm{y}}=-1.8 \mathrm{m} / \mathrm{s} \text { at } \mathrm{t} 1=10 \mathrm{s} \text { . }} \\ {\text { For the time interval from } \mathrm{t} 1=10 \mathrm{s} \text { to } \mathrm{t} 2=20 \mathrm{s} \text { the average acceleration of the dog has magnitude }} \\ {0.45 \mathrm{m} / \mathrm{s}^{\wedge} 2 \text { and direction } 31 \text { measured from the }+\mathrm{x} \text { -axis toward the }+\mathrm{y} \text { -axis. At } \mathrm{t} 2 \mathrm{z}=20 \mathrm{s} \text { , }}\end{array}
$$

$$
\begin{array}{l}{\text { (a) what are the } x \text { -and } y \text { -components of the dog's velocity? }} \\ {\text { (b) What are the magnitude and direction of the dog's velocity? }} \\ {\text { (c) Sketch the velocity vectors at } t 1 \text { and t2. How do these two vectors differ? }}\end{array}
$$

$$
v_{x}, v_{y} \ @ t_{1}=10 s
$$

$$
a_{ a v g}=0.45 \mathrm{m/s}^{2}
$$

$$
\theta=31^{\circ}
$$

$$
t_{2}=20 \mathrm{s} ? ?
$$

$$
V_{x}, v _y \quad ? ?
$$

$$
v , \theta \ ??
$$

$$
V x{2}=6.5 \mathrm{m} / \mathrm{s}
$$

$$
v_{y_{2}}=0.52 \mathrm{m/s}
$$

$$
\left(a_{avg}\right) [\begin{array}{l}{\ { a_{avg \  x }}} \\ {\ { a_{ avg \ y} }}\end{array}
$$

$$
a_{avg \ x}=a_{avg } \cos \theta=0.45 \cos (31)=0.39 \mathrm{m/s}^2
$$

$$
a_{avg \ y}=a_{avg} \sin \theta=0.45 \sin (31)=0.23 \mathrm{m} / \mathrm{s}^{2}
$$

$$
a_{avg-x }=\frac{\Delta V x}{\Delta t}=\frac{V_{x 2}-V x_{1}}{\Delta t}
$$

$$
\Delta V_{x}=a \Delta t
$$

$$
\left(v_{2}-v_{1}\right)=a \Delta t
$$

$$
V_{2}=V_{1}+a \Delta t
$$

$$
V_{x{2}}=V_{x{1}}+a_{\text {avg }{x}} \Delta t
$$

$$
=(2.6)+(0.39)(20-10)
$$

$$
\rightarrow V_{x{2}}=6.5 \mathrm{m/s}
$$

$$
a_{avg-y}=\frac{\Delta V y}{\Delta t}=\frac{V_{y_{2}}-V_{y_ 1}}{\Delta t}
$$

$$
v_{y_{2}}=v_{y_{1}}+a_{a v-y}(\Delta t)
$$

$$
=-1.8+(0.23)(20-10)
$$

$$
\rightarrow V_{ y_{2}}=0.52 \mathrm{m} / \mathrm{s}
$$

$$
V=\sqrt{V_{x}^{2}+V_{y}^{2}}
$$

$$
=\sqrt{(6.5)^{2}+(0.52)^{2}}=6.52 \mathrm{m} / \mathrm{s}
$$

$$
\theta=\tan ^{-1}\left(\frac{0.52}{6.5}\right)=4.6^{\circ}
$$

$$
\text { above the horrzontal }
$$

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