Answer to Question #183428 in Mechanics | Relativity for andrews

Question

2. 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

To be given in question

$\overrightarrow{r}$ =$(e^{-t}cost) i+(e^{-t}sin) j$

To be asked in question

Velocity=?

Phase =$\frac{3 \pi}{4}$

r.a=?

We know that

$r=(e^{-t}cost )i+(e^{-t}sint)j \rightarrow(1)$

Equation differenciate with respect to t

$v=\frac{dr}{dt} \rightarrow(2)$

Put value in eqution (1)and(2)

$v=\frac{d}{dt}((e^{-t}cost) i+(e^{-t}sint) j)$

$v=(-e^{-t}cost-e^{-t}sint)i-(e^{-t}sint+e^{-t}cost)j \rightarrow(3)$

$a=(2e^{-t}sint)i+(-2e^{-t}cost)j$

eqution (1) take mode

$\mid r\mid =$ $\sqrt{(e^{-t}cost)^2+(e^{-t}sint)^2}$

1. $r^2=e^{-2t}cos^2t+e^{-2t}sin^2t+2e^{-2t}costsint =2e^{-2t}$

$|v|=\sqrt{2e^{-2t}}=\sqrt2e^-t$

$cos(r.v)=\frac{r.v}{|r||v|}=\frac{-e^{-2t}}{e^-t\sqrt2e^{-t}}=-\frac{\sqrt2}{2}$

$$angle(r.v)=\frac{3\pi}{4}$$

(b)

$r.a=e^{-t}cost(2e^{-t}sint)+e^{-t}sint(-2e^{-t}cost)=0$