A Train is start from station with acceleration α after some time it retards with β and comes to rest at next station then calculate (1) accelerating time, (2) retarding time (3) maximum velocity ,(4)average velocity (5)distance between station. assumed (total time of journey =T)
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Secondary SchoolPhysics 8 points
A train starts from rest and accelerates uniformly at a rate of 2m/s^2 for 10s. It then maintains a constant speed for 200s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50s. Find : ( i ) the maximum velocity reached, ( ii ) the retardation in the last 50s, ( iii ) the total distance travelled, and ( iv ) the average velocity of the train.
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shubhamjoshi033
shubhamjoshi033 Virtuoso
i) Since the train accelerates for 10 sec and maintains a uniform speed after then until it starts retardation , hence the maximum speed will be at t = 10 sec
v = u + at
=> v = 0 + 2x10
=> Vmax = 20 m/s
ii) velocity before retardation = 20 m/s
velocity after retardation = 0
time = 50 sec
=> v = u + at
=> 0 = 20 + ax50
=> a = -20/50 = -0.4 m/s²
hence the retardation will be -0.4 m/s²
iii) Distance traveled in first 10 sec
s₁ = ut + at²/2
=> s₁ = 0 + 2x10²/2
=> s₁ = 100 m
distance traveled in the next 200 sec with a velocity of 20 m/s
s₂ = vt
= 20 x 200
= 4000 m
distance traveled in the next 50 sec
s₃ = ut + at²/2
=> s₃ = 20 x 50 - 0.4x50²/2
s₃ = 1000 - 500 = 500
hence total distance traveled
s = s₁ + s₂ + s₃ = 100 + 4000 + 500 = 4600 m
iv) Total distance = 4600
total time = 260 s
average velocity = total distance/total time
=> Vavg = 4600/260 = 17.7