# 6r question....

Discussion in 'MCAT: Medical College Admissions Test' started by Omyss, Aug 8, 2006.

1. ### Omyss Member 2+ Year Member

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for those who have taken 6r and still have the test can you clarify the question for me?

#28. The solutions say: Use Big Five #2, and call down the positive direction. Then v0 = v (its negative, because v0 is upward and were
calling down the positive direction) and vfinal = +v, so vfinal = v0 + at becomes v = (v) + gt, or 2v = gt, which gives t = 2v/g.
This means that t is inversely proportional to g. So, if g is decreased by a factor of 6, then t will increase by a factor of 6.

but how can you know that the initial velocity = - final velocty?

I used one of the other big 5 eq'ns:
d = VfT - 1/2at^2
thus since vf is zero t= square root of 2d/a
which gives me the anser that t would increease by a factor of root 6.

Why is this wrong?

2. ### MahSpoon is TOOOOO big 7+ Year Member

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the equation is X=ViT + 1/2at^2 (initial velocity) not final

I've never used a form with Vf and i substitute in -g as needed

The easiest way to approach this is to use
Vf = Vi -gT
When Vf is zero gT=Vi. Thus, if you have the same Vi and decrease g by 6, you must increase T by 6 to balance the equation. Hope this helps!

3. OP

### Omyss Member 2+ Year Member

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there are two equations that look similar but are slightly different.. and their both part of the big five:

d = VoT +1/2aT^2
and
d= VfT - 1/2aT^2 so why didn't this equation work.. thats weird.

4. ### IckeyShuffle MS1 t-minus 1.5 months.. 2+ Year Member

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im right there with ya bud. I put the exact same answer as you.

5. ### DiverDoc KCUMB 2012 10+ Year Member

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this is the equation you use. If i remember right, the answer is 50 meters or point B. Make sure you solve for T. Then use distance equals rate (given) multiplied by time.

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6. OP

### Omyss Member 2+ Year Member

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your're thinking of the wrong question lol... this one is asking by what factor the time changes...

7. ### xylem29 7+ Year Member

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use big 5 # 4 - d= v(initial)t + 1/2gt(sq)

to go up and come back down gives your displacement as zero. then just solve for t.

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