The correct option is B 0.1 ms−1
VB,A=10 ms−1
f=10 cm=0.1 m
u=1.9 m
As the focal length is positive, it is a concave mirror.
Now, 1f=1v+1u
1−0.1=1v+1−1.9
1v=11.9−10.1=181.9
∴v=−0.1055 m
From the mirror equation, we can write,
⇒f−1=u−1+v−1
Differentiating both sides w.r.t. time,
0=(−1)u−2dudt+(−1)v−2dvdt
⇒u−2dudt=−v−2dvdt
⇒1u2dudt=−1v2dvdt
Here, dudt→speed of object w.r.t. mirror
dvdt→speed of image w.r.t. mirror
Now, VB,A=dudt=40 ms−1
u=−1.9 m, v=−0.1055≈−0.1 m
∴1(−1.9)2×40=−1(−0.1)2×dvdt
After calculating, we get, dvdt=−0.11 m/s−1
Negative sign indicated that, as object is moving towards the mirror, the image is moving away from it.
Hence, (A) is the correct answer.