Answer to Question #236332 in Physics for eshiii

Question #236332

The position of a squirrel running in a park is given by π‘Ÿ = [(310.280π‘š/𝑠)𝑑 + (0.0360π‘š/𝑠 2 )𝑑 2 ]𝑖̂+ (0.0190π‘š/𝑠 3 )𝑑 3 𝑗̂. (a) What are 𝑣π‘₯(𝑑) and 𝑣𝑦(𝑑), the π‘₯- and 𝑦-components of the velocity of the squirrel, as functions of time? (b) At 𝑑 = 5.00 s, how far is the squirrel from its initial position? (c) At 𝑑 = 5.00 s, what are the magnitude and direction of the squirrel’s velocity?


1
Expert's answer
2021-09-20T10:01:21-0400

(a)

"v_x(t)=\\dfrac{dr_x}{dt} = (0.28\\ \\dfrac{m}{s})+(0.072\\ \\dfrac{m}{s^2})t,""v_y(t)=\\dfrac{dr_y}{dt} =(0.057\\ \\dfrac{m}{s^3})t^2."

(b) Let's first find "x"- and "y"-components of the final position of the squirrel at "t=5.0\\ s":


"r_x=(0.28\\ \\dfrac{m}{s})\\cdot5.0\\ s+0.036\\ \\dfrac{m}{s^2}\\cdot(5.0\\ s)^2=2.3\\ m,""r_y=(0.019\\ \\dfrac{m}{s^3})\\cdot(5.0\\ s)^3=2.4\\ m."

Finally, we can find the final position of the squirrel at "t=5.0\\ s" from the Pythagorean theorem:


"r=\\sqrt{r_x^2+r_y^2}=\\sqrt{(2.3\\ m)^2+(2.4\\ m)^2}=3.3\\ m."

(c) Let's first find "x"- and "y"-components of the squirrel's velocity at "t=5.0\\ s":


"v_x(t=5.0\\ s)= (0.28\\ \\dfrac{m}{s})+(0.072\\ \\dfrac{m}{s^2})\\cdot5.0\\ s=0.64\\ \\dfrac{m}{s},""v_y(t=5.0\\ s)=(0.057\\ \\dfrac{m}{s^3})\\cdot(5.0\\ s)^2=1.43\\ \\dfrac{m}{s}."

We can find the magnitude of the squirrel's velocity from the Pythagorean theorem:


"v=\\sqrt{v_x^2+v_y^2}=\\sqrt{(0.64\\ \\dfrac{m}{s})^2+(1.43\\ \\dfrac{m}{s})^2}=1.6\\ \\dfrac{m}{s}."

We can find the direction of the squirrel's velocity from the geometry:


"\\theta=tan^{-1}(\\dfrac{v_y}{v_x}),""\\theta=tan^{-1}(\\dfrac{1.43\\ \\dfrac{m}{s}}{0.64\\ \\dfrac{m}{s}})=66^{\\circ}."

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