train acquires the speed of 180 kilometre per hour in 25 seconds starting from rest at what distance does it rain cover in this time
Answers
Given that,
Acceleration a=−0.5m/s
2
Speed v=90km/h=25m/s
Using equation of motion,
v=u+at
Where,
v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the equation
Finally train will be rest so, final velocity,v=0
0=25−0.5t
25=0.5t
t=
0.5
25
t=50 sec
Again, using equation of motion,
S=ut+
2
1
at
2
Where, s = distance
v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the equation
Where S is distance travelled before stop
s=25×50−
2
1
×0.5×(50)
2
s=625 m
So, the train will go before it is brought to rest is 625 m.
Hence, A is correct.
Answer:
2250m
Explanation:
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.