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Motion in A Plane question

2015 · Q136
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Motion in A Plane question

2015 · Q136

NEETPhysicsMotion in A PlaneMCQ+4 / −1
A ship A is moving Westwards with a speed of 10 km h−-−1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h−-−1. The time after which the distance between them becomes shortest, is
  1. A
    525\sqrt 252​ h
  2. B
    10210\sqrt 2102​ h
  3. C
    0 h
  4. D
    5 h
View written solutionFree

Correct answer: D

V→A=10(−i^){\overrightarrow V _A} = 10\left( { - \widehat i} \right)VA​=10(−i)

V→B=10(j^){\overrightarrow V _B} = 10\left( {\widehat j} \right)VB​=10(j​)

V→BA=10j^+10i^=102 km/h{\overrightarrow V _{BA}} = {\rm{10}}\widehat {\rm{j}}{\rm{ + 10}}\widehat i{\rm{ = 10}}\sqrt 2 \,{\rm{km/h }}VBA​=10j​+10i=102​km/h

Distance OB = 100 cos 45° = 502\sqrt 2 2​ km

AIPMT 2015 Cancelled Paper Physics - Motion in a Plane Question 45 English Explanation

Time taken to reach the shortest distance between
A and B = OBVBA→=502102=5h{{OB} \over {\overrightarrow {{V_{BA}}} }} = {{50\sqrt 2 } \over {10\sqrt 2 }} = 5hVBA​​OB​=102​502​​=5h

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