A subway train starts from rest at a station and accelerates at a rate of 1.55 m/s^2 for 13.9 s. It runs at constant speed for 70.3 s and slows down at a rate of 3.51 m/s^2 until it stops at the next station.

Find the total distance covered!

Question

A subway train starts from rest at a station and accelerates at a rate of 1.55 m/s^2 for 13.9 s. It runs at constant speed for 70.3 s and slows down at a rate of 3.51 m/s^2 until it stops at the next station.

Find the total distance covered!

amistre64 · Answer

accelerates at a rate of 1.55 m/s^2 for 13.9 s.

a = 1.55
velocity=$\int$a = 1.55t
position=$\int$v = .725 t^2

    13.9
    13.9
-------
    1251
    417
  139
--------
   193.21
        .725
---------
      96605
    38642
135247
-----------
140.07725  meters so far, if i did it right

the speed hes going at this moment is:

v = 1.55
      13.9
      -----
      1395
      465
    155
   --------
    21.545  meters per sec.

It runs at constant speed for 70.3 s 

    21.545
        70.3
    -------
        6435
1508150
----------
1508.7935   meters more
+140.07725
------------
 1648.87075  meters total ... if im doing it right

and slows down at a rate of 3.51 m/s^2 until it stops at the next station. 

-3.51t  + 21.545 = 0 when t=21.545/3.51, gonna try a calculator on this one :)

      in 6.14 seconds or so

-1.755t^2 should give us the final stretch, and of course a calulator ...

1.755(6.14)^2 = 66.1628 meters more

 1648.87075
     66.162798
-------------
 1715.033548  meters 

maybe?

amistre64 · Answer

anyway to check it?

anonymous · Answer

You're 100% correct, and but I ended up with 149.738 for the first leg of the trip. Not sure how, going to have to go back and check my work!

amistre64 · Answer

thnx, i think the math tho is typoed in the middle1654.69075 is what the calulator tells me.. so im a little off i think

anonymous · Answer

Thank you, I am going to look over your work and see where I went wrong! Wonderful job!