of
A train , storting from rext picks up a ser
in lods. It continues to more at
the same speed for the next 2508.
then brought to rest in the next Sos. Plot
a speed time
A
It is
the train groph for the entire motion
Answers
Explanation:
answr
search
What would you like to ask?
11th
Physics
Motion in a Straight Line
Problems on Equation of Motion
A train starts from rest fr...
PHYSICS
A train starts from rest from a station with acceleration 0.2 m/s
2
on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s
2
. If total time spent is half an hour, then a distance between two stations is [ Neglect the length of the train]
MEDIUM
Share
Study later
CALCULATOR
Enter the know values to find unknown
Acceleration
Distance
Final Velocity
v = u + a*t
Initial Velocity u
m/s
Final Velocity v
m/s
Time t
s
Acceleration a
m/s²
RESET VALUES
ANSWER
Given,
Velocity v
1
achieve in time t
1
,due to acceleration, a
1
=0.2ms
−2
v
1
=u
1
+at
1
⇒v
1
=0+0.2t
1
Velocity reduce to zero in time t
2
, due to retardation, a
2
=−0.4ms
−2
v
2
=v
1
+at
2
⇒0=v
1
−0.4t
2
Add time
t
1
+t
2
=
0.2
v
1
+
0.4
v
1
⇒30×60=v
1
×7.5
⇒v
1
=240m/s
Apply kinematic equation
v
2
−u
2
=2as
s=
2a
v
2
−u
2
Total distance S
1
+S
2
=
2a
1
v
1
2
+
2a
2
v
1
2
=
2×0.2
240
2
+
2×0.4
240
2
=216km