Question A rocket is travelling along the path described by 2x) = 13. (1) with x and y in m. When it reaches point A where x 30 m, its speed is 66 m/s and it begins to accelerate along the path at a rate of mis? (where s is the arc length measured from Ainm) y = 60 Determine: a) [2] the exact (e. integers or rational numbers only) speed as a function of arc length s measured from point A mis b) [1] the exact (e integer or rational numbers only) radius of curvature as a function of x for x > 0 When the rocket reaches an x-coordinate of x = 70 m, it has traveled an arc length of approximately s = 41.57 mmeasured from point A (accurate to the number of sig digs displayed). Use this to determine... c) [1]... the tangential component of the acceleration at that instant: mi 0911 ... the normal component of the acceleration at that instant e) 10.5] .. the magnitude of the acceleration at that instant [1.5) ... the angle that acceleration vector makes in degrees measured clockwise relative to the positive x-axis: Check Last saved at 0.08:17 Previous page Finish attempt

(1)Given data,y=13 \ln \left(\frac{x}{30}\right)a, The Acceleration equation is,\begin{array}{l}a_{t}=\frac{95^{2}}{1250} \\V \frac{d v}{d s}=\frac{9 s^{2}}{1250} \\v d v=\frac{9 s^{2}}{1250} d s \\\int_{66}^{v} v d v=\frac{9}{1250} \int_{0}^{S} s^{r} d s \\\left.\left.\frac{V^{2}}{2}\right]_{66}^{V}=\frac{9}{1250} \frac{s^{3}}{3}\right]_{0}^{s} \\\frac{v^{2}}{2}-\frac{66^{2}}{2}=\frac{3 s^{3}}{1250} \\v^{2}-4356=\frac{65^{3}}{1250} \\V=\sqrt{0.00485^{3}+4356} \mathrm{~m} / \mathrm{s} \\\end{array}(2)b)we have the equation as,y(x)=13 \cdot \ln \left(\frac{x}{30}\right)\frac{d y}{d x}=13 \frac{1}{\frac{x}{30}}-\frac{1}{30}\frac{d y}{d x}=\frac{13}{x}\frac{d^{2} y}{d x^{2}}=\frac{-13}{x^{2}}we knew that,\begin{array}{l}r(x)=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2} \text {. }}{\frac{d^{2} y}{d x^{2}}} \\r(x)=\left[1+\left(\frac{13}{x}\right)^{2}\right]^{3 / 2} \\\frac{13}{x^{2}} \\r(x)=\frac{x^{2}}{13}\left[1+\frac{169}{x^{2}}\right]^{3 / 2} \\\end{array}(3)C,The tangential component of Acceleration is,\begin{aligned}a_{t} & =\frac{9 s^{2}}{250} \\& =\frac{9 \times 41.57^{2}}{1250} \quad S=41.57 \text { (given } \\& =\frac{15552.58}{1250} \\\therefore \quad & 12.44 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}dConsider,\begin{aligned}V & =\sqrt{0.0048 \mathrm{~s}^{3}+4356} \\& =\sqrt{0.0048(41.57)^{3}+4356} \\& =\sqrt{24774092.4} \\V & =4977.35 \mathrm{~m} / \mathrm{s}\end{aligned}(7)\begin{aligned}V & =\sqrt{4700.81116} \\\therefore V & =68.56 \mathrm{~m} / \mathrm{s}\end{aligned}Since, r(x)=\frac{x^{2}}{13}\left[1+\frac{169}{x^{2}}\right]^{3 / 2}\begin{aligned}\text { At } x & =70 \mathrm{~m} \\r(70) & =\frac{70^{2}}{13}\left[1+\frac{169}{70^{2}}\right]^{3 / 2} \\& =1633.33[1.0344]^{3 / 2} \\& =1633.33(1.052) \\r(70) & =1718.26 \mathrm{~m}\end{aligned}The normal component of acceleration is,\begin{aligned}a_{n} & =\frac{v^{2}}{r(70)} \\& =\frac{68.56^{2}}{1718.56} \\& \therefore a_{n}=2.7351 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}As per guidelines we are allowed to answer only first 4 subparts of the given question.Please kindly post the other question separately ...