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# Answer to Question #171430 in Mechanical Engineering for edjee

Question #171430

A Stone has an initial velocity of 200 ft per see up to the right at a slope of 4 to 3. The components of acceleration are constant at a_x=-12 ft per second² and a_y=-20 ft per second². Compute the radius of curvature at the start and at the top of the path

1
2021-04-01T01:04:32-0400
"a_x=-12, a_y=-20"

"\\vec r''(t)=-12\\vec i-20\\vec i"

"\\vec v_0=200\\cdot\\dfrac{3}{\\sqrt{3^2+4^2}}\\vec i+200\\cdot\\dfrac{4}{\\sqrt{3^2+4^2}}\\vec j"

"\\vec v_0=120\\vec i+160\\vec j"

"\\vec r'(t)=(120-12t)\\vec i+(160-20t)\\vec j"

"\\vec r'(t)\\times \\vec r''(t)=\\begin{vmatrix}\n \\vec i & \\vec j & \\vec k \\\\\n 120-12t & 160-20t & 0 \\\\\n -12t & -20t & 0 \n\\end{vmatrix}"

"=\\vec k((120-12t)(-20t)-(160-20t)(-12t))"

"=-480t\\vec k"

"||\\vec r'(t)\\times \\vec r''(t)||=480|t|"

"||\\vec r'(t)||=\\sqrt{(120-12t)^2+(160-20t)^2}"

"=4\\sqrt{34t^2-580t+2500}"

"K(t)=\\dfrac{||\\vec r'(t)\\times \\vec r''(t)||}{||\\vec r'(t)||^3}"

"=\\dfrac{480t}{64(34t^2-580t+2500)^{3\/2}}"

"=\\dfrac{15t}{2(34t^2-580t+2500)^{3\/2}}"

"R(t)=\\dfrac{2(34t^2-580t+2500)^{3\/2}}{15t}"

At the start of the path: "t=0"

"R(0)=\\infin"

At the top of the path: "v_y=0"

"160-20t=0"

"t=8"

"R(8)=\\dfrac{2(34(8)^2-580(8)+2500)^{3\/2}}{15(8)}=3.6 (ft)"

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