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$$
\begin{array}{l}{\text { A squirrel has } x \text { -and } y \text { -coordinates }(1.1 m, 3.4 m) \text { at time } t 1=0 \text { and }} \\ {\text { coordinates }(5.3 m,-0.5 m) \text { at time } t 2=3 s \text { For this time interval, }} \\ {\text { find (a) the components of the average velocity, }} \\ {\text { (b) the magnitude and direction of the average velocity. }}\end{array}
$$

$$
r_{1}=x_{1} \hat i+y_{1} \hat{j}=1.1 \hat{i}+3.4 \hat{j} \quad m
$$

$$
r_{2}=x_{2} \hat{\imath}+y_{2} \hat{\jmath}=5.3 \hat{\imath}-0.5 \hat{\jmath} \quad \mathrm{m}
$$

$$
V_{x_{\text {avg }}} ? \quad V_{y_{\text {avg} }} ? ?
$$

\(∵ V_{a vg}=\frac{d}{t} \)

$$
\rightarrow V_{avg}{x}=\frac{d x}{t} \quad, \quad V_{avg } y=\frac{dy}{t}
$$

$$
\rightarrow V_{a v g} x=\frac{5.3-1.1}{3} \quad, \quad V_{a v g} y=\frac{-0.5-3.4}{3}
$$

$$
=1.4 \mathrm{m} / \mathrm{s} \quad \quad \ =-1.3 \mathrm{m} / \mathrm{s}
$$

$$
\text { (b) } V_{\text {avg }} \text { ? }
$$

$$
V_{a v g}=\sqrt{V_{a vg} {x}^{2}+V_{a vg} y^{2}}
$$

$$
=\sqrt{(1.4)^{2}+(-1.3)^{2}}
$$

$$
=1.9 \mathrm{m} / \mathrm{s}
$$

$$
\tan \alpha = \frac{(V_{ avg})_ y}{(V_{ a vg})_{x}}=\frac{-1.3}{1.4}=-0.9286
$$

$$
\rightarrow \alpha=360^{\circ}
$$

$$
\begin{array}{l}{\text { The position of a squirrel running in a park is given by } r=\left[(0.28 \mathrm{m} / \mathrm{s}) \mathrm{t}+\left(0.036 \mathrm{m} / \mathrm{s}^{\wedge} 2\right) \mathrm{t}^{\wedge} 2\right] 1+\left(0.019 \mathrm{m} / \mathrm{s}^{\wedge} 3\right) \mathrm{t}^{\wedge} 3 \mathrm{j}} \\ {\text { (a) } \mathrm{What} \text { are } \mathrm{V} x(t) \text { and } \mathrm{V} y(\mathrm{t}), \text { the } x \text { -and } y \text { -components of the velocity of the squirrel, as functions of time? }(\mathrm{b}) \mathrm{At}} \\ {\text { t-5 } \mathrm{s} \text { , how far is the squirrel from its initial position? (c) } \mathrm{At} \text { the } 5 \mathrm{s} \text { what are the magnitude and direction of the }} \\ {\text { squirrel's velocity? }}\end{array}
$$

$$
r=\left[0.28 t+0.036 t^{2}\right] i+0.019 t^{3} \hat{j}
$$

$$
t=5
$$

$$
V_{x}(t), V_{y}(t) ? ?
$$

$$
x, y \ @ t=5s
$$

$$
V , \theta \ @ t=5s
$$

$$
V=\frac{d r}{d t}
$$

$$
V_{x}=\frac{d v{x}}{d t}, \quad V_{y}=\frac{d r y}{d t}
$$

$$
=\frac{dr x}{d t}=\frac{d\left(0.28 t+0.036 t^{2}\right)}{d t}
$$

$$
=0.28+(0.036)(2) t=0.28+0.072 t
$$

$$
=\frac{dr y}{d t}=\frac{d\left(0.019 t^{3}\right)}{d t}=3(0.019) t^{2}=0.057 t^{2}
$$

$$
V_{x}=0.28+0.072 t
$$

$$
V_{y}=0.057 t^{2}
$$

$$
r_{x}=0.28 t+0.036 t^{2}
$$

$$
r_ y=0.019 t^{3}
$$

$$
\longrightarrow \quad t=5
$$

$$
\left.\begin{array}{l}{r_{x}=2.3 m} \\ {r_ y=2.375 m}\end{array}\right\} \begin{array}{l}{x} \\ {y}\end{array}
$$

$$
V_{x}=0.28+0.072 t \rightarrow @ \ t=5 s \rightarrow V_{x}=0.64 \mathrm{m} / \mathrm{s}
$$

$$
V_ y=0.057 t^{2} \longrightarrow @ \ {t}=5s \longrightarrow V_ y=1.425 \mathrm{m} / \mathrm{s}
$$

$$
\rightarrow V=\sqrt{{V_ x}^{2}+{V_ y}^{2}}=\sqrt{(0.64)^{2}+(1.425)^{2}}=1.56 \mathrm{m} / \mathrm{s}
$$

$$
\text { at } t=5 s \rightarrow V_{x}=0.64 \mathrm{m} /
$$

$$
V_ y=1.425 \mathrm{m}
$$

$$
\tan \theta=\frac{V_ y}{V_ x}
$$

$$
\rightarrow \theta=\tan ^{-1} \frac{V_ y}{V_{x}}
$$

$$
=\tan ^{-1} \frac{1.425}{0.64}=65.8^{\circ}
$$

$$
(\text { c.c.w from }+x \text { -axis })
$$

$$
\begin{array}{l}{\text { A web page designer creates an animation in which a dot on a computer screen has a position of }} \\ {\left.r=\left[4 \mathrm{cm}+2.5 \mathrm{cm} / \mathrm{s}^{\wedge} 2\right] \mathrm{t}^{\wedge} 2\right] 1+(5 \mathrm{cm} / \mathrm{s}) \mathrm{t} \mathrm{j}} \\ {\text { (a) Find the magnitude and direction of the dot's average velocity between } \mathrm{t}=0 \text { and } \mathrm{t}=2 \mathrm{s}} \\ {\text { (b) Find the magnitude and direction of the instantaneous velocity at } \mathrm{t}=0, \mathrm{t}=1 \mathrm{s}, \text { and } \mathrm{t}=2 \mathrm{s}}\end{array}
$$

$$
V_\text { av ?? } \quad  \theta { ?? }
$$

$$
V \quad @ \quad t=0, \ t=1s, \ t=2s
$$

$$
\vec{r}=[(4+(2.5) {t}^{2}] \hat{\imath}+(5 t) \hat{\jmath}
$$

$$
At \  t=0 \quad \longrightarrow \vec{r}=4 \hat{i}
$$

$$
\text { At } t=2 s \rightarrow \vec{r}=[(4+(2.5)(2)^{2}] \hat{i}+(5(s)] \hat J
$$

$$
=14 \hat{i}+10 \hat{j}
$$

$$
(V_{avg} )_ x=\frac{\Delta x}{\Delta t}=\frac{14-4}{2-0}=\frac{10}{2}=5 \mathrm{cm} / \mathrm{s}
$$

$$
\left(V_{\text {avg }}\right)_ y=\frac{\Delta y}{\Delta t}=\frac{10-0}{2-0}=\frac{10}{2}=5 \mathrm{cm} / \mathrm{s}
$$

$$
\left(V_{\text {avg }}\right)=\sqrt{({{V_{arg}})_{x}}^{2}+({{V_{avg}})_{y}}^{2}}=\sqrt{(5)^{2}+(5)^{2}}=7.1 \mathrm{cm} / \mathrm{s}
$$

$$
\tan \alpha=\frac{\left(V_{avg}\right)_ y}{\left(V_{avg}\right)_{x}}
$$

$$
=\frac{5}{5}=1
$$

$$
\rightarrow \theta=\tan ^{-1}(1)
$$

$$
=45^{0}
$$

$$
\vec{V}=\frac{\vec{dr}}{d t}
$$

$$
=\frac{d\left[\left(4+2 \cdot 5 t^{2}\right) \hat{i}+(5t) \hat{j}\right]}{d t}
$$

$$
=2(2.5) t+5=(5 t) \hat i+5  \hat j
$$

$$
\left.\begin{array}{rl}{a t \ t=0 \rightarrow V_{x}=} & {5 t=5(0)=0} \\ {V_ y=5} & {\text { cm/s }}\end{array}\right] \rightarrow V=\sqrt{(0)^{2}+(5)^{2}}=5 \mathrm{cm} / \mathrm{s}
$$

$$
\begin{array}{c}{a t \  t=1 \longrightarrow V_{x}=5 t=5(1)=5 \mathrm{cm} / \mathrm{s}} \\ {V_ y=5 \mathrm{cm} / \mathrm{s}}\end{array}] \rightarrow V=\sqrt{5^{2}+5^{2}}=7.1 \mathrm{cm} / \mathrm{s}
$$

$$
\left.\begin{array}{c}{\text { at } t=2 \rightarrow V_{x}=5 t=5(2)=10 \mathrm{cm} / \mathrm{s}} \\ {V_{y}=5 \mathrm{cm} / \mathrm{s}}\end{array}\right] \rightarrow \mathrm{V}=\sqrt{10^{2}+5^{2}}=11 \mathrm{cm} / \mathrm{s}
$$

$$
\theta=\tan ^{-1}\left(\frac{5}{0}\right)=90^{\circ}\leftarrow v=\sqrt{(0)^{2}+(5)^{2}}=5 \mathrm{cm} / \mathrm{s}
$$

$$
\theta=\tan ^{-1}\left(\frac{5}{5}\right)=45^{\circ}\leftarrow \quad V=\sqrt{5^{2}+5^{2}}=7.1 \mathrm{cm} / \mathrm{s}
$$

$$
\theta=\tan ^{-1}\left(\frac{5}{10}\right)=27^{\circ}\leftarrow V=\sqrt{10^{2}+5^{2}}=11 \mathrm{cm} / \mathrm{s}
$$

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