a) The maximum velocity of the train is 14 m/s
b) Distance:
first stage:
t1= 30 s
V0=0 m/s
ΔV=14 m/s
a=ΔV/t1
s1=V0*t1+0,5*a*t1²
s1=0,5*(ΔV/t1)*t1²
s1=0,5*ΔV*t1
second stage:
t2=110
Δt= t2- t1 =100 - 30 = 70 s
V=s/t
s=V*t
s2=ΔV*(t2-t1)
third stage:
V0=14 m/s
Vk=0
ΔV=14 m/s
t3=120
Δt=t3-t2 =120-100 = 20 s
s3=V0*Δt-0,5*(ΔV/Δt)*Δt²
s3= 0,5*V0*Δt
s3=0,5*V0*(t3-t2)
s=0,5*ΔV*t1+ ΔV*(t2-t1) + 0,5*V0*(t3-t2)
s=0,5*14*30 + 14*(100-30) + 0,5*14*(120-100)
s=1330 m
c)
tc=120 s
Vavg=s/tc
Vavg=1330/120
Vavg ≈ 11,08 m/s
d)
Initial acceleration:
t1= 30 s
V0=0 m/s
Vk=14 m/s
ΔV=14 m/s
a=ΔV/t1
a=14/30 = 7/15 ≈ 0,47 m/s²
e)
Greater is deceleration on the braking (when the train enters the station)
Δt= 120-100 = 20 s
V0=14 m/s
Vk=0 m/s
ΔV=14 m/s
a=ΔV/t1
a=14/20 = 7/10 ≈ 0,7 m/s²
The inertia force is greater for higher accelerations/decleration.
