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^{th }of September 2021 from 12:30 Hrs to 13:30 Hrs as per the Indian Standard Time. The site may not be accessible during the maintenance hour.Object and mirror move towards each other with velocity 2m/s and 3m/s respectively. What is the velocity of the image with respect to the ground?

@swati The key to this simple problem is the mirror equation that relates the object distance to the image distance.

Although the question does not explicitly state it, we can assume the following:

The object is moving towards the mirror at a velocity $2\ \frac{m}{s}$ with respect to the ground and the mirror is moving towards the object at a velocity $3\ \frac{m}{s}$.

When all distances are measured from the centre of the mirror, let object distance be $u$ and image distance be $v$. We also have to follow the sign convention. However, to provide an easy and intuitive answer, let is also assume that the mirror is a plane mirror.

**For plane mirrors:**

It is well established that $v=u$ (ignoring the sign).

Thus the time rate of change of $v$ must be the same as the time rate of change of $u$.

$\frac{dv}{dt}=\frac{du}{dt}$.

To correctly find the value of $\frac{du}{dt}$, we have to consider the mirror to be at rest with us. Hence the object velocity now becomes relative to the mirror. Since the object and the mirror are moving towards each other, the magnitudes must be added.

Then, with respect to the mirror (and thus to us),

$\frac{dv}{dt}=\frac{du}{dt}=2+3=5\ \frac{m}{s}$.

This means that the image is also moving towards the mirror at a velocity $5\ \frac{m}{s}$, with respect to the mirror. Thus the image velocity with respect to the ground is $8\ \frac{m}{s}$. (Answer)

**Steps for spherical mirrors:**

- Use the mirror equation: $ \frac{1}{u}+ \frac{1}{v}= \frac{1}{f}$.
- Use coordinate sign convention.
- Assign positive directions along the axis of the mirror.
- Find the object and image distances and velocities WRT the mirror first.
- Add the image velocity WRT the mirror to the mirror velocity WRT the ground to find the image velocity WRT the ground.

@thephysicist I have understood ^_^

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