Dismiss Notice
Check out the new Application Assistant, where you can calculate your LizzyM score, see how you rank compared to other applicants, and see a list of schools where similar students were accepted.

linear motion Q

Discussion in 'MCAT Study Question Q&A' started by Addallat, 09.26.14.

  1. SDN is made possible through member donations, sponsorships, and our volunteers. Learn about SDN's nonprofit mission.
  1. Addallat

    Addallat 5+ Year Member

    Joined:
    06.02.10
    Messages:
    126
    Status:
    Medical Student
    A particle moving at 5 m/s reverses its direction in 1 s to move at 5 m/s in the opposite direction. If acceleration is constant, what distance does it travel?

    A. 1.25 m
    B. 2.5 m
    C. 5 m
    D. 10 m

    Answer is B. 2.5 m

    I can solve this for displacement real easy using the following equation for uniform accelerated motion x=Vi(t)+1/2(a)(t^2) (answer would be zero), but I have no clue how to approach solving this for distance.

    The best I could come up with:

    Distance = Average Speed * time

    initial speed is = 5 m/s

    final speed is = 5 m/s

    Time = 1/2 second? why???

    Average speed = (initial speed+final speed)/2
    10/2 = 5 m/s

    Distance = 5 m/s * 1/2 second = 2.5 meters
    ^^^^^^^^^^^^^^^^^^
    WHY ARE THEY TAKING TIME as = 1/2 SECOND, so confused as to why time would be 1/2 second
     
    Last edited: 09.26.14
  2. SDN Members don't see this ad. About the ads.
  3. Hadi7183

    Hadi7183 2+ Year Member

    Joined:
    12.08.13
    Messages:
    63
    Location:
    NWT
    Status:
    Pre-Medical
    The question is asking for distance not displacement. It can approached like a free falling object problem. At t=1/2, the object comes to full stop and changes its direction. All you need to do is find displacement at t=1/2 and multiply it by 2. This gives you the distance (2.5 m), but the displacement it zero.
     
  4. popopopop

    popopopop 2+ Year Member

    Joined:
    12.18.11
    Messages:
    986
    Location:
    DFW/Houston
    Status:
    Medical Student (Accepted)
    The particle decelerates from 5 m/s to 0 in 1 second. It's final velocity is not 5 m/s, that's how I'm solving it.

    Vf = 0 m/s and Vi = 5 m/s. My Vavg. = (5+0)/2 = 2.5 m/s.

    v = d/t, 2.5=d/1 sec = 2.5 meters traveled.
     
  5. Addallat

    Addallat 5+ Year Member

    Joined:
    06.02.10
    Messages:
    126
    Status:
    Medical Student
    "
    The particle decelerates from 5 m/s to 0 in 1 second. It's final velocity is not 5 m/s, that's how I'm solving it.

    Vf = 0 m/s and Vi = 5 m/s. My Vavg. = (5+0)/2 = 2.5 m/s.

    v = d/t, 2.5=d/1 sec = 2.5 meters traveled.
    "



    oh man makes so much sense thank you!
     

About the ads

Share This Page