a train start from rest and accelerates uniformly at 100m - minute ^_2 for 10 minutes. It then , maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded and comes to rest in the 5minutes. Draw a velocity - time graph and use it to find: (¡) the maximum velocity reached , (¡¡) the retardation in the last minutes, (¡¡¡) today distance travelled, and (¡v) the average velocity of the train
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in which class are you
I am in 9th
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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 Distance
=4600/260=17.69 m/s
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