Question

An image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 25/3 m to 50/7 in 30 seconds. What is the speed of the object in km per hour?

Answer 1

We need to find the corresponding object distances of the two image distances. Then we need to find the speed by using the difference in object distances divided by the time.

Formula used: In this solution we will be using the following formulae;
$f = \dfrac{R}{2}$ where $f$ is the focal length and $R$ is the radius of curvature,$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ where $f$ is the focal length of a mirror, $u$ is the object position from the mirror, and $v$ is the image position from the mirror.
$s = \dfrac{d}{t}$ where $s$ is the speed of an object, $d$ is the change in position and $t$ is the time taken for the object to change position.

Complete step-by-step answer:
To calculate the speed, we need to calculate the object positions. We can use the mirror equation, as in
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ where $f$ is the focal length of a mirror, $u$ is the object position from the mirror, and $v$ is the image position from the mirror
And $f = \dfrac{R}{2}$ where$R$ is the radius of curvature.
Hence, $f = \dfrac{{20}}{2} = 10m$
By inserting all known values into the mirror equation, for the initial position, we have
$- \dfrac{1}{{10}} = - \dfrac{1}{{\dfrac{{25}}{3}}} + \dfrac{1}{u}$ (since the focal length and image position of a convex mirror by convention are regarded as negative).
Then
$\dfrac{1}{u} = - \dfrac{1}{{10}} + \dfrac{3}{{25}} = \dfrac{{ - 5 + 6}}{{50}} = \dfrac{1}{{50}}$
$\Rightarrow u = 50m$
Similarly for the final position, we have
$\dfrac{1}{{{u_2}}} = - \dfrac{1}{{10}} + \dfrac{7}{{50}} = \dfrac{2}{{50}}$
$\Rightarrow {u_2} = 25m$
Hence, the distance travelled is
$d = {u_2} - u = 25 - 50 = - 25m$
Hence, speed is
$s = \dfrac{d}{t} = \dfrac{{25}}{{30}} = 0.833m/s$ or $3km/h$

Note: For clarity, we neglect the negative sign in the distance because speed is not a vector and hence the direction of the object should not be considered. The negative sign only signifies that the object reduces its distance to the mirror in time.