An observer is moving along the line AB as
shown. When image of object O is first visible
to observer, he starts from rest with the
acceleration of 2cm/sec, then the time for
which image is visible to observer is
Answers
The time for which image is visible to observer is 9 seconds.
Explanation:
From ΔOPQ
tanθ = 27 / 2 x 10 ---(1)
From ΔQAC
tanθ = x / 20
=> x / 20 27/ 2 (10) => x = 27
Now the distance for which image will be visible.
AB = 2x + 27 = 81 cm
So for time using second equation of motion.
S = ut + 1/2 (2)t^2
81 = 1/2 (2)t^2
t = 9 seconds
Thus the time for which image is visible to observer is 9 seconds.
Also learn more
A man standing on a platform observes that frequency of sound of whistle emitted by train drops by 140 Hz if the velocity of sound in air is 330 M per second and the speed of the train is 70 M per second the frequency of the whistle is ?
https://brainly.in/question/6714950
Answer: time t = 9 seconds
Explanation:
In the ΔOPQ we get
tanθ = 27 / 2 x 10 ---(1)
In the ΔQAC
tanθ = x / 20
that is => x / 20 = 27/ 2 (10) => x = 27
The distance for which image will be visible.
AB = 2x + 27 = 81 cm
To find the time we are using second equation of motion.
S = ut + 1/2 (2)t^2
81 = 1/2 (2)t^2
t = 9 seconds