# linear motion Q

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

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1. ### Addallat 5+ Year Member

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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

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
3. ### Hadi7183 2+ Year Member

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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 2+ Year Member

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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 5+ Year Member

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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!