Answer: Option D
:
D
Given:
Initial object distance, $u=-60m$
Speed of object, $v_o=20m{s}^{-1}$
Focal length, $f=20m$
Time interval,$t=2s$
Applying mirror formula to find the initial position of the image $v$:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}+\frac{1}{-60}=\frac{1}{20}$
$v=15m$
Distance travelled by the object is the product of speed and time.
$s_o=v_{o}t$
$s_o=20\times 2=40m$
Therefore, now the object distance:
$u_f=u+s_o=-60+40=-20m$
Applying mirror formula to find the final position of the image $v_f$:
$\frac{1}{v_f}+\frac{1}{u_f}=\frac{1}{f}$
$\frac{1}{v_f}+\frac{1}{-20}=\frac{1}{20}$
$v_f=10m$
Average speed is the ratio of total distance travelled to the total time taken.
Averagespeed of the image:
$v_i=\frac{v-v_f}{t}$
$v_i=\frac{15-10}{2}=2.5m{s}^{-1}$