Writing correct sign of quantities

[Rhdjfgjgj]

Homework Statement

My doubt is regarding the signs for the given picture .
Here I didn't understand how my sir wrote dy/dt=vi-3vo.
I felt I should have been
3vo -vi because this vector is opposite to the velocity vector of the object . This is a question from geometrical optics.

Relevant Equations

No equations needed

 

[BvU]

Hi,

All very interesting, but what's this all about ? What is the complete problem statement ?
Which way is positive ?

$\$

 

[Lnewqban]

Here I didn't understand how my sir wrote dy/dt=vi-3vo.
I felt I should have been
3vo -vi because this vector is opposite to the velocity vector of the object .

Respect to the ground, which value of velocity is greater, the velocity of the lens or mirror (3Vo) or the velocity of the image projected by it (Vi)?

Which subtraction gives a positive dy/dt?

Note that distance y grows positive (to the right) from the mirror or lens, while distance x grows positive (to the left) from the mirror or lens.

 

[Rhdjfgjgj]

My sir taught us to find velocity of image when mirror is at rest. So we used relative motion ka concept. Wrt to mirror, object is moving with 4vo away from the mirror along normal. He assumed that the image is formed at i and assumed that it's moving with vi as shown .

He assumed object distance as x and image distance as y.
Now just see the expression of dx/dt and dy/dt . The expression is written wrt mirror
Since vi is opposite of vo dy/dt has to be -(vi-3vo) . Like almost all these question types he used same thing only .
vi is velocity of image wrt ground.

Respect to the ground, which value of velocity is greater, the velocity of the lens or mirror (3Vo) or the velocity of the image projected by it (Vi)?

Which subtraction gives a positive dy/dt?

Note that distance y grows positive (to the right) from the mirror or lens, while distance x grows positive (to the left) from the mirror or lens.

Since y is increasing vi must be grater than 3vo. And since he wrote dx/dt as 4vo I assume he must have taken the left direction as + for velocity.

Hi,

All very interesting, but what's this all about ? What is the complete problem statement ?
Which way is positive ?

$\$

 

[Rhdjfgjgj]

Friends im waiting for your reply

 

[vela]

In terms of relative velocities, you have $\vec v_{IG} = \vec v_{IM} + \vec v_{MG}$ where
\begin{align*}
\vec v_{IG} &= \text{velocity of the image with respect to the ground} = v_I \,\hat i \\
\vec v_{IM} &= \text{velocity of the image with respect to the mirror} = \frac{dy}{dt} \hat i\\
\vec v_{MG} &= \text{velocity of the mirror with respect to the ground} = 3v_o\,\hat i
\end{align*} If you solve that equation for $dy/dt$, you get what your instructor wrote down.

More intuitively, if the image moves to the right (holding the mirror position fixed), $y$ increases, so $dy/dt$ and $v_I$ have the same sign. If the mirror moves to the right (with the image position fixed), it will cause $y$ to decrease, so $3v_o$ comes in with a negative sign.

 

[Rhdjfgjgj]

In terms of relative velocities, you have $\vec v_{IG} = \vec v_{IM} + \vec v_{MG}$ where
\begin{align*}
\vec v_{IG} &= \text{velocity of the image with respect to the ground} = v_I \,\hat i \\
\vec v_{IM} &= \text{velocity of the image with respect to the mirror} = \frac{dy}{dt} \hat i\\
\vec v_{MG} &= \text{velocity of the mirror with respect to the ground} = 3v_o\,\hat i
\end{align*} If you solve that equation for $dy/dt$, you get what your instructor wrote down.

More intuitively, if the image moves to the right (holding the mirror position fixed), $y$ increases, so $dy/dt$ and $v_I$ have the same sign. If the mirror moves to the right (with the image position fixed), it will cause $y$ to decrease, so $3v_o$ comes in with a negative sign.

Well, in that case vo(object velocity) should have been -4voi

 

[vela]

It seems $x$ is a distance, not a position, and $v_o$ is a speed, not a velocity.