Question

a subway train starts from rest at a station and accelerates at a rate of 3 metre per second square for 10 seconds it then runs at constant speed for 22nd and decelerate at 5 metre per second square until it stops at the next station the distance between the two station is ​

Answer 1

Answer:

Step-by-step 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=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=22.4*6.4-(1/2)3.5*6.4*6.4=143.36-71.68=71.68

Total distance = s1+s2+s3=1796.48m