train starting from rest is accelerated by 0.5 m./s2 for 10s, then it moves with constant speed for the next 10 s and finally, retarding at a rate of 1.0 m /s2 , it stops. (i) Draw timespeed graph of the train. (ii) Calculate, from the graph, the total distance travelled, the average speed during first 20 seconds and the maximum speed of the train.
Answers
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.