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Q.

Three elephants A, B and C are moving along a straight line with constant speed in same direction as shown in figure. Speed of A is 5m/s and speed of C is 10m/s. Initially separation between A and B is ‘d’ and between B and C is also d. When ‘B’ catches ‘C’ separation between A and C becomes 3d. Then the speed of B will be

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a

7.5 m/s

b

20 m/s

c

15 m/s

d

5 m/s

answer is A.

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Detailed Solution

$V_{A C} = V_{A} - V_{C} = 5 - 10 = - 5 m / s V_{B C} = V_{B} - V_{C} = V_{B} - 10 I n i t i a l s e p a r a t i o n b e t w e e n ‘ A ’ a n d ‘ C ’ = 2 d D i s \tan c e b e t w e e n ‘ A ’ a n d ‘ C ’ = 3 d ∴ S_{A B} = 3 d - 2 d = d t = \frac{d}{V_{B C}} = \frac{d}{V_{A C}} V_{B} = 15 m / s$

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