Two cars A and B are moving with same speed of 45 km/h along the same direction. If a third car C coming from opposite direction with a speed of 36 km/h meets twocars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)
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Answer:
V A =V B =45km/hr
V C =36km/hr
Let x is the distance between A and B.
Velocity of approach between A and B.
⇒V AC =V A +V B =(45+36)km/hr=81km/hr
V BC =V B+V C =(45+36)km/hr=81km/hr
⇒ Velocity of approach =
Time difference in meeting A and B/Distance between A and B
⇒△t=5min= 5 /60
⇒81= x×60/5
⇒x=5×81/60
=6.75km
Hence, the answer is 6.75km.
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