A subway train starts from rest at a station and accelerates at a rate of 1.60m/s^2 for 14.0s. It runs at constant

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station-acceleraates

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A subway 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.

Answer:

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=(at^2)/2=1.614*14/2=156.8

Distance during the next 70 seconds,
s2=22.4*70=1568 (using the formula s=ut+(1/2)at^2)

Time taken to stop, use the equation v=u+at
0=22.4-3.5t
t=6.4 seconds

Distance travelled during these 6.4seconds,
s3=22.46.4-(1/2)3.56.4*6.4=143.36-71.68=71.68

Total distance = s1+s2+s3=1796.48m