A train starts from rest from a station and travels with uniform acceleration of 0.5m/s for 20 seconds. it travels with uniform velocity for another 30 seconds. the btakes are then applied so that a uniform retardation is obtained and the train comes to rest in further 10seconds. Scetch the velocity time graph of this motion using your graph, calculate the total distance travelled by the train
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Answer:
This is the correct answer with proper explanation........ PLEASE FOLLOW
Explanation:
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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