1 Answer
Solution :-
Given a man 160cm tall walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1m/sec
To find the rate at which the length of his shadow increases when he is 1m away from the pole
Let AB be the lamp post and let MN be the man of height 160cm or 1.6m.
Let AL = l meter and MS be the shadow of the man
Let length of the shadow MS = s (as shown in the below figure)
Given as the man walks at the speed of 1.1m/sec
So, \( \dfrac{dl}{dt}\)= 1.1m/sec ...(i)
Therefore, the rate at which the length of the man's shadow increases will be \( \dfrac{ds}{dt}\)
Considering ΔASB,
Then considering ΔMSN,
Therefore, from equation (ii) and (iii)
1= 2.75s
By applying derivative with respect to time on both sides
Thus, the rate at which the length of his shadow increases by 0.4 m/sec, and it is independent to the current distance of the man from the base of the light.
Answered by Sakshi | 1 year agoRelated Questions
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