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The distance traveled can be obtained from the area of the time-velocity graph. As we know when the train passes the post it will travel 100m in 10s. In the graph this region is a trapezoid and its area will be 100.
Area of trapezoid `= (11+V)/2xx10`
`(11+V)/2xx10 = 100`
`V = 9`
So the velocity of train at the start of post is 9m/s.
Using gradient of the graph we can say;
`V/T = 11/(T+10)`
`9/T = 11/(T+10)`
`T = 45s`
So the front of train will pass B after 45 seconds it started travel and with a velocity of 9m/s.
Posted by jeew-m on July 3, 2013 at 1:31 PM (Answer #1)
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