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bject l0UR IinoDerive the second equation of motion. A train is travelling at a speed of 90 km/h. Brakesare applied so as to produce a uniform acceleration of -0.5 m/s. Find how far the trainwill go before it is brought to rest.15. |
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Answer» Given v = 0u= 90km/h = 25m/sa=-0.5m/s^2s=? v^2-u^2=2ass=v^2-u^2/2as=0^2-25^2/2×-0.5s=0-625/2×-0.5s=-625/-1s= 625m Ans) The train will need a distance of 625m to stop. Distance= average velocity ×time s =( v+u/2)t s= (u+at+u/2)t s= 2u+at×t/2 s= 2u+at^2/2 s= ut+at^2/2 s= ut+1/2at^2 |
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