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A particle starts from rest, accelerates at 2 m/then goes for constant speed for 30s and thenat 4 m/s till it stops. What is the distance travelledIDCE 2001: AIIMS 2002.0ay 750 m(b) 800 m(c) 700 m(d) 850 m |
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Answer» option 1st is correct For first 10 sec distance s1 = 1/2 at^2 1/2×2×10×10 = 100mAnd velocity reached= v =u+atv= 0+2×10 = 20m/sFor next 30ss2= 20× 30 = 600m Then under retardation of 4m/s^2v^2 = 2as320× 20 = 2×4×s3S3 = 50m Total s1+ s2+ s3 100+600+50 = 750m |
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