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A ship A is moving Westwards with a spee...

A ship A is moving Westwards with a speed of 10 km/h and a ship B 100 km South of A is moving northwards with a speed of 10 km/h . The time after which the distance between them becomes shortest is

A

0 h

B

5 h

C

`5 sqrt(2)`h

D

`10 sqrt(2)`h

Text Solution

Verified by Experts

The correct Answer is:
B
Let the required time be t hours.
If we suppose the x-axis along eastward direction and the y-axis along northward direction , then the position of ship A after time t,
`vecr_A=(-10hati)t`
The position of ship B after time t,
`vec_B=-100hatj+(10hatj)t`
Therefore, position of B with respect to A,
`vec_B-vec_A =(10t)hati+(10 t-100)hatj`
so, `|vec_B-vec_A| =sqrt((10t)^2+(10t-100)^2)`
`=sqrt(100t^2+100t^2-2000t+10000)`
`=10 sqrt(2) sqrt(t^2-10t+50)`
This distance becomes the shortest when `(t^2-10t+50)` becomes minimum, i,e.,
`d/(dt) (t^2-10t+50) =0 or, 2t -10 =0`
`therefore t=5 hours`
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