11. A train starts from rest with a constant acceleration
3 m s-2 for 5 s. Now it moves with a constant speed for
the next 300 s and finally retards to come to rest again
in the next 15 seconds.
(i) What is the maximum speed of the train?
(ii) What is the retardation produced in the train?
(iii) What is the total distance moved by the train?
(iv) What is the average velocity of the train?
Answers
Answered by
2
Answer:
Time=Distance/Speed
Answered by
6
Answer:
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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