Asked by parul
a train starts from rest and accelerates uniformly at 100m/min^2 for 10 min.then it maintains a constant velocity for remaining 20 min.brakes are applied and train uniformly retards and comes to rest in next 5 min.
a)the maximum velocity reached
b)the retardation in last 5 min
c)total distance traveled
d)the average velocity of the train.
a)the maximum velocity reached
b)the retardation in last 5 min
c)total distance traveled
d)the average velocity of the train.
Answers
Answered by
Henry
a. V = Vo + at
V = 0 + 100m/min^2 * 10min = 1000 m/min
b. a=(V-Vo)/t=(0-1000) / 5=-200m/min^2
c. d1 = 0.5a*t^2 = 50*10^2 = 5000 m.
d2 = 1000m/min * 20min = 20,000 m.
d3 = Vo*t + 0.5a*t^2
d3 = 1000*5 + (-100)*5^2 = 2500 m.
d1+d2+d3 = 5000 + 20000 + 2500=27,500 m
= Tot. dist. traveled.
d. T = t1 + t2 + t3 = 10+20+5 = 35 Min.
Va=(d1+d2+d3)/T=27,500m/35min=786 m/min
= Average velocity.
V = 0 + 100m/min^2 * 10min = 1000 m/min
b. a=(V-Vo)/t=(0-1000) / 5=-200m/min^2
c. d1 = 0.5a*t^2 = 50*10^2 = 5000 m.
d2 = 1000m/min * 20min = 20,000 m.
d3 = Vo*t + 0.5a*t^2
d3 = 1000*5 + (-100)*5^2 = 2500 m.
d1+d2+d3 = 5000 + 20000 + 2500=27,500 m
= Tot. dist. traveled.
d. T = t1 + t2 + t3 = 10+20+5 = 35 Min.
Va=(d1+d2+d3)/T=27,500m/35min=786 m/min
= Average velocity.
Answered by
Anonymous
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Answered by
saleem
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