Calculating angle between velocity and position vector

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

I took $\frac { d\vec r}{dt} = 0$ . This gives $t_m = \frac { 1} {\sqrt{ 2}}$ ...(1)

To calculate $\alpha$ ,

$\cos{ \alpha } = \frac { \vec v \cdot \vec r}{|\vec v||\vec r|}$ ...(2)

Then, magnitude of $\vec v$ is 0 at t = $t_m$.

So, R.H.S. of (2) becomes $\frac { 0}0$ . So, $\alpha$ could be anything between $- \pi$ to $\pi$ .Another approach was to put $\cos{ \alpha } = \frac { \vec v \cdot \vec r}{|\vec v||\vec r|}$ as a function of t and then to reduce it and then solve it at $t= t_m$. But, this led me into difficult calculation yielding again $\frac { 0}0$. I don’t know why this approach should give a different answer, I just thought of it as some times, while solving for limit ,some factor gets canceled leading answer to non $\frac { 0}0$ form.

 

[TSny]

I took $\frac { d\vec r}{dt} = 0$ .

Does this condition correspond to the particle being at its maximum distance from the origin?

 

[Pushoam]

Does this condition correspond to the particle being at its maximum distance from the origin?

Sorry, What I did was: $\frac{ds}{dt} = 0$ where s is the magnitude of displacement.

 

[TSny]

Sorry, What I did was: $\frac{ds}{dt} = 0$ where s is the magnitude of displacement.

OK

Then, magnitude of $\vec v$ is 0 at t = $t_m$.

How did you arrive at this statement?

 

[Pushoam]

OK
How did you arrive at this statement?

Sorry,
I calculated $\frac { ds} {dt} = 0$ .

Then, I got the thought that, for the distance between the origin and the system should be 0, $\frac { d\vec r}{dt} = 0$, which is wrong.

I did not remain careful about the fact that $\frac { ds} {dt}$ and $\frac { d\vec r}{dt}$ represent two different physical quantities, that they mean different. This carelessness was the mistake.

Identified with the above thought, I took it for granted that $|\vec v|= 0$ at t = $t_m$ .

$|\vec v| = \sqrt{ t^2 – t + \frac 1 {4 t}} e^{-t}$

So, at t = $t_m$ , $\vec v \neq \vec 0, |\vec v|\neq 0$

$\vec v \cdot \vec r = 0$

So, at $t = t_m , \alpha = \frac { \pi}2$

So, the answer is option (d).

Is this correct?So, at t = $t_m$ , $\vec v \neq \vec 0, |\vec v|\neq 0$

$\vec v \cdot \vec r = 0$

So, at $t = t_m , \alpha = \frac { \pi}2$

So, the answer is option (d).

Is this correct?

 

[TSny]

Looks good.

 

[Pushoam]

Thanks.