An electric train moves from rest with a uniform acceleration of 1.5m/s square for the first ten seconds and continue to accelerate at 0.5 m/s square for a further 20s,it continue at constant speed for 90s and finally take 30s to decellerate uniforming to rest.
(a) draw a graph of speed against time for the journey (b) from your graph or otherwise deduce the total distance travelled (c) what is the average speed of the whole journey
Answers
Answer:
ANSWER
Train starts from rest, hence the initial velocity u = 0.
It moves with acceleration = 2m/s2 for half minute (30 seconds).
Distance covered in this time interval is given by:
S=ut+½at
2
=0+½×2×30×30
=900m
Velocity attained by this acceleration after 30 seconds:
v=u+at
=>v=0+2x30
=>v=60m/s
From this velocity, brakes are applied and train comes to rest in 60 seconds.
The retardation is given by:
v=u–at
=>0=60–a×60
=>a=1m/s
2
Distance covered in this time:
$$V2= u2 + 2aS$$
=>0=(60)2+2(−1)S
=>0=3600–2S
=>S=3600/2=1800m.
So, total distance moved =900m+1800m=2700m.
Maximum speed of the train=60m/s.
Position of the train at half its maximum speed.
Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s.
(I) When the train is accelerating with an acceleration of 2 m/s,
time at which speed = 30m/s is:
v=u+at
=>30=0+2xt
=>t=15s
At 15s, distance covered from origin is:
S=ut+½at
2
=0+½×2×15×15
=225m
(II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is:
v=u–at
=>30=60–1xt
=>t=30s
At 30s, distance covered is:
S=ut–½at
2
=60x30–½x1x(30)2
=1800–(15x30)
=1800–450
=1350m (from the initial 900m covered).
So, distance from origin =900+1350m=2250m.
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