a train starts from rest and accelerates uniformly at 100 min-2 for 10 min .It then maintains a constant velocity for 20 minutes.the brakes are then applied,and the train is uniformly retarded.it comes to rest in 5 min. draw a velocity time graph find the maximum velocity reached, the retardation in the last 5min. , the total distance traveled and the average velocity ms-² taking=10ms-²
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Answer:
Correct option is
C
17.69ms
−1
Total distance covered(s) = Distance during acceleration(s
1
) + distance during uniform motion(s
2
) + distance during retardation(s
3
)
For acceleration,
v=u+a∗t
v
max
=2×10=20 m/s
s=ut+(1/2)at
2
s
1
=(1/2)×2×10
2
=100 m
For uniform motion,
s=vt
s
2
=20×200
=4000 m
During retardation,
v=u+at
0=20+50t
=−0.4 m/s
2
s
3
=ut+(1/2)at
2
=20×50+(1/2)×(−0.4)×50
2
=500 m
Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m
Total time taken, t=t
1
+t
2
+t
3
=260 s
Average velocity, v
avg
=
Total Time
Total DDistan =4600/260=17.69 m/s
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