Answer to Question #183469 in Mechanics | Relativity for frank

Question

. A particle moves so that its position vector at time t is given by ˜ r = e −t cost ˜ i + e −t sin t ˜ j. Show that at any time t, (a) its velocity ˜ v is inclined to the vector ˜ r at a constant angle 3π/4 radians. (b) its acceleration vector is at right angles to the vector ˜ r.

Answer 1

Explanations & Calculations

• Once you know the position vector as a function of time (t), the velocity & the acceleration vectors can be found by derivating the position vector once & twice respectively.
• Useful operator is

$\qquad\qquad \begin{aligned} \small \frac{d(u.v)}{dt}&=\small \frac{du}{dt},v+u.\frac{dv}{dt} \end{aligned}$

• And once you know any two vectors, the included angle can be found by performing the dot product of vectors.
• And I hope you know how to perform the dot product for 2 vectors [ex: $\small (p\it{i}+q\it{j})$ and $\small (r\it{i}+s\it{j})$ ]

$\qquad\qquad \begin{aligned} \small \underline{a} .\underline{b}&=\small |\underline{a}||\underline{b}|\cos\theta\\ \small \cos\theta&=\small \frac{\underline{a}.\underline{b}}{|\underline{a}||\underline{b}|}\cdots\cdots(1) \end{aligned}$

• Then following what is mentioned above in the calculations ahead, needed quantities can be found.

$\qquad\qquad \begin{aligned} \small \underline{r}&=\small e^{-t}\cos\theta.\it{i}+e^{-t}\sin\theta.\it{j}\cdots\cdots(given)\\\\ \small \underline{v}&=\small \frac{dr}{dt}\\ \small&=\small (-e^{-t}\cos\theta-e^{-t}\sin\theta)\it{i}+(-e^{-t}\sin\theta+e^{-t}\cos\theta)\it{j}\\ &=\small -e^{-t}(\cos\theta+\sin\theta)\it{i}-e^{-t}(\sin\theta-\cos\theta)\it{j}\\\\ Then\\ \small \underline{r}.\underline{v}&=\small -e^{-2t}(\cos^2\theta+\sin\theta \cos\theta )-e^{-2t}(\sin^2\theta-\sin\theta\cos\theta)\\ &=\small -e^{-2t}(\cos^2\theta+\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta)\\ &=\small -e^{-2t}(1)\\ &=\small -e^{-2t}\\\\ \small |\underline{r}|&=\small \sqrt{(e^{-t}\cos\theta)^2+(e^{-t}\sin\theta)^2}\\ &=\small \sqrt{(e^{-t})^2\times(\cos^2\theta+\sin^2\theta)}\\ &=\small e^{-t}\\\\ \small |\underline{v}|&=\small \sqrt{[-e^{-t}(\cos\theta+\sin\theta)]^2+[-e^{-t}(\sin\theta-\cos\theta)]^2}\\ &=\small \sqrt{(-e^{-t})^2\times[(\cos\theta+\sin\theta)^2+(\sin\theta-\cos\theta)^2 ]}\\ &=\small e^{-t}.\sqrt{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta}\\ &=\small e^{-t}.\sqrt{1+1}\\ &=\small \sqrt2 e^{-t}\\\\ \small \cos\theta&=\small \frac{-e^{2t}}{e^{-t}\times\sqrt2e^{-t}}\\ &=\small -\frac{1}{\sqrt2}\\ \small \theta&=\small \frac{3\pi}{4}\cdots\cdots(\because\,\pi\leq \theta\leq0) \end{aligned}$

• Therefore, for the first case, the two vectors make this angle between them.

• Then for the acceleration,

$\qquad\qquad \begin{aligned} \small \underline{a}&=\small \frac{d^2r}{dt^2}= \frac{dv}{dt}\\ &=\small (e^{-t}(\cos\theta+\sin\theta)-e^{-t}(-\sin\theta+\cos\theta))\it{i}-(-e^{-t}(\sin\theta-\cos\theta)+e^{-t}(\cos\theta+\sin\theta))\it{j}\\ &=\small 2e^{-t}\sin\theta .\it{i}-2 e^{-t}\cos\theta.\it{j}\\\\ Then\\ \small \underline{a}.\underline{r}&=\small (2e^{-t}\sin\theta\times e^{-t}\cos\theta)+(-2e^{-t}\cos\theta\times e^{-t}\sin\theta)\\ &=\small 2e^{-2t}\sin\theta\cos\theta-2e^{-2t}\sin\theta\cos\theta\\ &=\small 0 \end{aligned}$

• When the dot product becomes zero for two vectors, that means (according to [1],) the cosine of the contained angle is zero hence they are 90 degrees apart which places them in an orthogonal position.
• Then the acceleration vector is perpendicular to the position vector.