Asubway train starts from rest at a station and accelerates at a rate of 1.60m/s^2 for 14.0s. it runs at constant speed for 70.0s and slows down at a rate of 3.50m/s^2 until it stops at the next station. find the total distance covered.

1796.48 m

Explanation:

Velocity after 14 seconds can be given by the equation,

Final Velocity=Initial Velocity + Acceleration*Time

v=0+1.6×14=22.4m/s^2

Distance covered during these 14 seconds,

s1=(a×t^2)/2=1.6×14×14/2=156.8

Distance during the next 70 seconds,

s2=22.4×70=1568 (using the formula s=u×t+(1/2)at^2)

Time taken to stop, use the equation v=u+at

0=22.4-3.5×t

t=6.4 seconds

Distance traveled during these 6.4seconds,

s3=22.4×6.4-(1/2)3.5×6.4×6.4=143.36-71.68=71.68

Total distance = s1+s2+s3=1796.48m

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