Physics MCAT Assessment #22. Kinematics

[LuminousTruth]

"If a ball is thrown upward vertically with velocity v and returns to original height at time t, what would happen to t if g was reduced to g/6?

A) Increase by a factor of 6
B) Increase by a factor of 6^(1/2)
C) Decrease by a factor of 6
D) Decrease by a factor of 6^(1/2)

I chose: "B"
The correct answer was: "A"

-----------------------------------------------------------------------------

I used d=vi*t + 1/2at^2 but the solutions used V=at.

You are given initial velocity v, distance (2d), acceleration (g/6), and you are looking for t. You are not given final velocity so we can use eq d=vi*t + 1/2at^2. (from TPR Big 5). Why wouldn't d=1/2at^2 get the correct answer?

---

Members don't see this ad.

 

[shouldvestudied]

You're problem is that the distance isn't 2d. This is projectile motion where the object is shot straight up and then returns to its original position. Therefore the distance is equal to 0.

Thus:

0 = vt - (1/2)at^2
vt = (1/2)at^2
v = (1/2)at
2v/a = t

plug in a/6 for a to get:

12v/a = t = 6(2v/a)

So time increases by a factor 6 when g is decreased by a factor of six.

 

[Trayshawn]

Another way to do it:

you DO know the final velocity. Its the negative of the initial velocity since we know it returns to the original height
so..

v final = v initial + gt

[-v initial] = v initial +gt

-(2v initial)/g = t

now plug in (g/6) for g and you get..

-(2v initial)/(g/6) = 6*[-(2v initial)/g] = t (the new time)

thus the new time equals 6t (6 times the old time)