A train start from rest from a station with acceleration 0.2m/s2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4m/s2. if total time spent is half an hour, then distance between two station is
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Distance will be 216km
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Answer:
216 m
Explanation:
Given,
Velocity v1 achieved in time t1 due to acceleration,
a1 = 0.2 m/s^2
v1 = u1 + at1
=> v1 = 0 + 0.2 t1
Velocity reduces to zero in time t2 due to retardation,
a2 = -0.4 m/s^2
v2 = v1 + at2
=> 0 = v1 - 0.4 t2
Add time
t1 + t2 = (v1/0.2) + (v1/0.4)
=> 30 X 60 = v1 X 7.5
Note: 1/2 hour means 30 X 60 seconds
=> v1 = 240 m/s
Apply kinematic equation v^2 - u^2 = 2as
s = (v^2 - u^2)/ 2a
Total distance
s1 + s2 = (v1^2/2a1) + (v1^2/2a2) = 240^2/ (2 X 0.2)
= 216 km
Please mark this answer as the 'Brainliest'
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