A train starts from rest and accelerates uniformly at a rate of 2 ms −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 displacement
Answers
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 m/s.
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Answer:
4600 m,please mark this as the brainliest answer!!please!
Explanation:
S₁=ut +1/2*(at²)=0+2*10²/2=100m
v=u+at=0+2*10=20m/sec
therefore
S₂=20*200=4000m
finally
v=u +at
hence 0=20-a*50
a=20/50=0.4
so S₃= 20*50-0.4*50²/2
=1000-500=500m
total=s₁+s₂+s₃=4600m
i hope it helps you!