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A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s2 for 14.0s. It runs constant speed for 70.0s and slows down at a rate of 3.50 m/s2 until it stops at the next station find the total distance covered.
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The total displacement is the sum of d1+s2+d3, where d1d1=.5*1.6*14^2 we also need v1=1.6*14 d2=v1*70 d3=v1*t2-.5*3.5*t2^2 where 0=v1-3.5*t2 d1=156.8 m v1= 22.4 m/s d2=1568 m t2=6.4 seconds d3=71.68 m total distance is 1.8 km
i think he's right
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