A trsin starting from 1 station acceleration uniformly over a distance of 5km moves with a constant speed 35km retarded uniformly over the last 10km coming to a stop at another station 50k. Away from the 1st ststion draw velocity time graph of the train
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A train start from rest from a station and travels with uniform acceleration 0.5ms^-2 for 20s it travelswithwwith uniform velocity for 30s d brakes are then applied so that a uniform retardation is obtained and the train comes to rest in a further 10s sketch the velocity time graph of the motion using your graph calculate the total distance traveled by the train
Answer by Alan3354(59731) (Show Source): You can put this solution on YOUR website!
A train start from rest from a station and travels with uniform acceleration 0.5ms^-2 for 20s it travels with uniform velocity for 30s d brakes are then applied so that a uniform retardation is obtained and the train comes to rest in a further 10s sketch the velocity time graph of the motion using your graph calculate the total distance traveled by the train
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vt graph, t = x-axis, v = y-axis
Start at the Origin.
The 1st part is y = 0.5x from x = 0 to x = 20
The speed after 20 seconds = 0.5*20 = 10 m/sec
The end-point is (20,10)
s = at^2/2 --> distance = 200 meters
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The 2nd part is y = 10 from x = 10 to x = 40
--> end-point is (40,10)
Total distance = 200 + 300 = 500 meters
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The brakes change its speed from 10 m/sec to 0 in 10 seconds --> a = 1 m/sec/sec
Its average speed for the 10 seconds is 5 m/sec
s = 5*10 = 50 meters
Total distance = 550 meters.
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email via the TY note and I'll send the graph.
Answer by Alan3354(59731) (Show Source): You can put this solution on YOUR website!
A train start from rest from a station and travels with uniform acceleration 0.5ms^-2 for 20s it travels with uniform velocity for 30s d brakes are then applied so that a uniform retardation is obtained and the train comes to rest in a further 10s sketch the velocity time graph of the motion using your graph calculate the total distance traveled by the train
-------------
vt graph, t = x-axis, v = y-axis
Start at the Origin.
The 1st part is y = 0.5x from x = 0 to x = 20
The speed after 20 seconds = 0.5*20 = 10 m/sec
The end-point is (20,10)
s = at^2/2 --> distance = 200 meters
-----
The 2nd part is y = 10 from x = 10 to x = 40
--> end-point is (40,10)
Total distance = 200 + 300 = 500 meters
---
The brakes change its speed from 10 m/sec to 0 in 10 seconds --> a = 1 m/sec/sec
Its average speed for the 10 seconds is 5 m/sec
s = 5*10 = 50 meters
Total distance = 550 meters.
=================
email via the TY note and I'll send the graph.
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Answer:
velocity time graph
Explanation:
t=X axis , v=Y axis
y
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