Question
Download Solution PDFA बिंदूपासून 34 m/s वेगाने सुरू होतो. 5 सेकंदांनंतर, B त्याच बिंदूपासून 44 मीटर/से वेगाने A चा पाठलाग सुरू करतो. A ने B पकडण्यापूर्वी B ने प्रवास केलेले अंतर किती असेल?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिले:
A ची गती = 34 मी/से
बी चा वेग = ४४ मी/से
वापरलेले सूत्र:
एका दिशेने सापेक्ष गती = गती1 -वेग2
गणना:
⇒ A ने 5 सेकंदात कापलेले अंतर = 34 x 5 = 170 मी
⇒ B wrt A चा सापेक्ष वेग = (44 - 34) m/s = 10 m/s
आता, A ला विश्रांतीसाठी घेऊन, A पकडण्यासाठी B ला 170 मीटर अंतर कापावे लागेल
⇒ A पकडण्यासाठी B ने लागणारा वेळ = 170 / 10 = 17 सेकंद
⇒ एकूण अंतर B = 17 x 44 = 748 मी
⇒ म्हणून, योग्य पर्याय 1 आहे
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