Question

Calculate the magnitude of velocity of the image $\left(V_s\right)$ with respect to a stationary observer in the following situation ( Object moving at an angle with the mirror and mirror is stationary).

• A. $5\sqrt{2}m/s$
• B. $5\sqrt{5}m/s$
• C. $5\sqrt{10}m/s$
• D. $5\sqrt{6}m/s$

Answer 1

The correct option is A $5\sqrt{2}m/s$

Let's resolve the velocity of the object as we discussed in the hint.

Velocity of object along the direction perpendicular to the mirror (horizontal direction),
$\underset{v_o\left(hor\right)}{\overset{}{\to }}=5m/s$ (right)
Therefore, horizontal velocity of the image,
$\underset{v_s\left(hor\right)}{\overset{}{\to }}=5m/s$ (left)

Now, the vertical velocity of the image is equal to that of the object, as required from object distance = image distance.
We have,
As, $\underset{v_o\left(ver\right)}{\overset{}{\to }}=5m/s$ (down)
Therefore, $\underset{v_s\left(ver\right)}{\overset{}{\to }}=5m/s$ (down)
Therefore, total velocity, $\overrightarrow{V_s}$, has a magnitude of $\sqrt{(5m/s{)}^2+(5m/s{)}^2}=5\sqrt{2}\frac{m}{s}.$