jennykim3
Answered

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a subway train starts from rest at a station and accelerates uniformly at 1.8m/s^2 for 15 sec. then it runs at a constant speed for the next 35 sec. before decelerating at 2.0m/s^2 until it comes to rest at the next station. FIND THE DISTANCE between the stations and the AVERAGE SPEED of the train.
(please show step by step and the formula that is used. Thank you!)

A Subway Train Starts From Rest At A Station And Accelerates Uniformly At 18ms2 For 15 Sec Then It Runs At A Constant Speed For The Next 35 Sec Before Decelerat class=

Sagot :

AL2006
This looks so scary just because there are so many parts to it.
Take them one at a time:

First piece:
From rest, accelerates 1.8 m/s² for 15 sec.
Speed at the end of the 15 sec = 1.8 x 15 = 27 m/s
Distance covered = (average speed) x (time) = (13.5) x (15) = 202.5 meters

Middle piece:
Starts this piece at 27 m/s.
Runs at constant speed for 35 sec.
Covers (27 x 35) = 945 meters
Finishes this piece (15 + 35) = 50 seconds after leaving the first station
Finishes this piece (202.5 + 945) = 1147.5 meters from the first station.

Last piece:
Starts this piece 50 seconds after leaving the first station,
         1147.5 meters from it, still moving at 27 m/s.
Decelerates at 2.0 m/s² .
--> It takes 27/2.0 = 13.5 seconds to come to a stop.
--> Distance covered during this piece = (average speed) x (time)
           = (13.5 m/s) x (13.5 sec) = 182.25 meters
Finishes this piece (50 + 13.5) = 63.5 seconds after leaving the first station
Finishes this piece (1147.5 + 182.25) = 1,329.75 m from the first station  

Distance between the stations (along the track) = 1,329.75 meters.

Average speed = (total distance) / (total time) = 1,329.75 / 63.5 = 20.941 m/s.

I sure hope I didn't make any stupid arithmetic blunders here,
because I really don't feel like going back and checking it.
Let me know, and I'll fix them.


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