Question

A particle moves along the curve y = x^3/2 such that its distance from the origin, measured along the curve, is given by s = t^3 . Determine the acceleration in vector form when t = 2 seconds. The units are inches and seconds.

Answer 1

$Let\:displacement\:vector\:be\:D=xi+yj\\ i\:and\:j\:are\:perpendicular\:unit\:directional\:vectors\:\:\:\:\:and\:y=\frac{x^3}{2}\\\\ D=xi+\frac{x^3}{2}j\\ \\ S=\sqrt{x^2+y^2}=\sqrt{x^2+\frac{x^6}{4}}=t^3\\\\ at\:t=2s\:\:\:\:\:\:\:\sqrt{x^2+\frac{x^6}{4}}=2^3\\\\ x^2+\frac{x^6}{4}=64\\\\ x=2.478\\ \\ x^2+\frac{x^6}{4}= t^6\\ \\ Taking\:derivative\\ \\ 2x\frac{\mathrm{d} x}{\mathrm{d} t}+6\frac{x^5}{4}\frac{\mathrm{d} x}{\mathrm{d} t}=6t^5\\\\ \frac{\mathrm{d} x}{\mathrm{d} t}=1.323\:in/s$

$\frac{\mathrm{d} x}{\mathrm{d} t}(2x+3x^5/2)=6t^5\\ Taking\:Derivative\:again\\\\ \frac{\mathrm{d}^2 x }{\mathrm{d} t^2}(2x+3x^5/2)+(\frac{\mathrm{d} x}{\mathrm{d} t})^2*(2+15x^4/2)=30t^4\\\\ \frac{\mathrm{d}^2 x }{\mathrm{d} t^2}=-0.127in/s^2 \\\\ \frac{\mathrm{d}^2 D}{\mathrm{d} t^2}=\frac{\mathrm{d}^2 x }{\mathrm{d} t^2}i+(\frac{\mathrm{d}^2 x }{\mathrm{d} t^2}*1.5*x^2+(\frac{\mathrm{d} x}{\mathrm{d} t})^2*3*x)j=-0.127in/s^2\:i+11.48j$