Two stations A and B are 10 km apart in a straight track, and a train
starts from A and comes to rest at B. For three quarters of the distance, the
train is uniformly accelerated and for the remainder distance uniformly
retarded. If it takes 15 minutes over the whole journey, find its acceleration,
retardation and the maximum speed it attains.
Answers
Acceleration is about 0.033 m/s and retardation is about 0.099 m/s
and the maximum speed it attains ≈ 22.22 m/s ( 80 km/hr )
V² - U² = 2aS
a = a₁ , U = 0 starts from stop S = (3/4) 10km = 7500 m
V² - 0 = 2a₁(7500)
=> V² = 15000a₁
Final Velocity = 0 as stop , retardation a₂ and S = 10000 - 7500 = 2500 m
0² - V² = 2(-a₂)(2500) a₂ represent magnitude only
=> V² = 5000a₂
Equating Both
15000a₁ = 5000a₂
=> 3a₁ = a₂
V = U + at
V = 0 + a₁t₁
V = a₁t₁
0 = V + (-a₂)t₂
=> 0 = a₁t₁ + (-a₂)t₂
=> a₁t₁ = a₂t₂
=> a₁t₁ = 3a₁t₂
=> t₁ = 3t₂
t₁ + t₂ = 15
3t₂ + t₂ = 15
=> 4t₂ = 15
=> t₂ = 15/4 mins
t₂ = 225 secs
t₁ = 3t₂ = 675 secs
V² = 15000a₁
Substitute V = a₁t₁
(a₁t₁)² = 15000a₁
=> a₁ ≈ 0.033 m/s
=> a₂ ≈ 0.099 m/s
V = a₁t₁ ≈ 22.22 m/s = 80 km/hr
Acceleration is about 0.033 m/s and retardation is about 0.099 m/s
and the maximum speed it attains ≈ 22.22 m/s ( 80 km/hr )