A person of height 1 . 6 m is walking away from a lamp post of height 4 m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60 cm s − 1 , the speed of the tip of the person’s shadow on the ground with respect to the person is _____ cm s − 1 .

Step-by-step solution

Given: Speed of person = d x 1 d t = 60   cm   s - 1 Also Speed of tip of person's shadow = d x 2 d t . As ∆ A B E and ∆ D C E are similar triangles, therefore we can write 4 x 2 = 1 . 6 x 2 - x 1 ⇒ 4 x 2 - 4 x 1 = 1 . 6 x 2 ⇒ 2 . 4 x 2 = 4 x 1 Differentiate both sides w.r.t. t , we get 2 . 4 d x 2 d t = 4 d x 1 d t ⇒ d x 2 d t = 4 2 . 4 60 = 100   cm   s - 1 Now, speed of the tip of the person's shadow on the ground with respect to the person will be v &#85