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Two skaters have weight in the ratio 4:5...

Two skaters have weight in the ratio `4:5` and are 9m apart, on a smooth frictionless surface. They pull on a rope stretched between them. The ratio of the distance covered by them when they meet each other will be

A

`5:4`

B

`4:5`

C

`25:16`

D

`16:25`

Text Solution

Verified by Experts

The correct Answer is:
A
Acceleration of the skaters will be in the ratio
`(F)/(4):(F)/(5) or 5:4`
Now according to the problem, `s=0+1/2at^(2)` we get
`(s_1)/(s_2)=(a_1)/(a_2) = 5/4`.
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