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Question 1158496: A Train on a straight railway track starts from rest and accelerates at 2.0 m/s2 in 10sec. Then the train covers a distance of 300 m at constant speed for 20 sec until the brakes are applied, stopping the train in a uniform manner in an additional 5.0 s. (a) How far is the train in first 30sec? (b) What is the average velocity of the train for the complete journey?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! a)
v=u+at
v= 0+2*10
v= 20m/s...........................(1)
s=ut+1/2 * a*t^2
=1/2* 2*100
= 100 m
b) 300 m for 20 s
v= 15m/s
deceleration distance = distance = 0.5 x deceleration x time^2 = 0.5 x 2.0 m/s^2 x (5 s)^2 = 25 m
Average velocity = (100+300+25 )=425/35 ( 35 s total time taken)
= 12.14 m/s
Answer by ikleyn(52765)  (Show Source):
You can put this solution on YOUR website! .
A Train on a straight railway track starts from rest and accelerates at 2.0 m/s2 in 10 sec.
Then the train covers a distance of 300 m at constant speed for 20 sec until the brakes
are applied, stopping the train in a uniform manner in an additional 5.0 s.
(a) How far is the train in first 30sec?
(b) What is the average velocity of the train for the complete journey?
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The problem is posed INCORRECTLY: its different parts are not consistent and contradict each other.
For example, it says " Then the train covers a distance of 300 m at constant speed for 20 sec ".
But after 10 seconds, the train has the speed v = at = 2*10 = 20 m/s,
therefore, in the next 20 seconds, if it moves at the constant speed,
it covers 20*20 = 400 meters, not 300 meters, as the problem states in the post.
Therefore, the problem can not be solved under these self-contradictory conditions.
As posted, it is only good to throw it to the closest garbage bin. Or to re-edit it accordingly.
There is no need to explain that obvious fact, that under these circumstances,
the "solution", given by the other person, makes no sense, too.
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