Kinematics (from AAMC 6)

[Tokspor]

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Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time t. If the value of g were reduced to g/6 (as on the moon), then t would:

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 understand that you can use the equation v(f) = v(i) + a*t to get the answer A. But why can't you use y = v(i) + (1/2)*g*t^2?

If we consider the peak in the second half of the arc of the projectile to be the starting point, v(i) is 0 at that peak.

So y = (1/2)*g*t^2, but we have to keep in mind that t is actually 1/2 the total travel time now, since we're starting at the second half of the projectile.

Solving for t, we set t = sqrt[(2*y)/g]. Knowing at if we multiply t by 2, we get the total traveling time, the total traveling time should equal 2*sqrt[(2*y)/g]. But if we divide g by 6, we still have to multiply t by 6^(1/2). This indicates B to be the answer. What is wrong with my reasoning here?

 

[Isoprop]

y is not the same in both equations. when g changes, y changes too.
Also, your equation is wrong. it's y = v(i)t + (1/2)*g*t^2

Edit: actually, I meant to say that y at its maximum height will change as g changes, but y does not change when the ball drops to the same height. In that case, you get y = 0.

So, 0 = vt + 1/2gt^2
vt = -1/2gt^2
-2v/g = t

Reducing g by a factor of 6 will increase t by a factor of 6. You get the same answer.

 

[jerrywheat]

that doesn't make sense. how can vt = -1/2gt^2. Since the ball always has an acceleration, there is no initial velocity which is what the v stands for. what i mean is that vt=0...... thats actually a really hard question. id be mad if they put that on the mcat. if you want to know how to do it, ask.

 

[zmatrix]

that doesn't make sense. how can vt = -1/2gt^2. Since the ball always has an acceleration, there is no initial velocity which is what the v stands for. what i mean is that vt=0...... thats actually a really hard question. id be mad if they put that on the mcat. if you want to know how to do it, ask.

It's y(t) = v(i)t + (1/2)*g*t^2, y (height) changes as time passes.

Therefore vt = -1/2gt^2 is only true when y(t)=0

 

[Isoprop]

that doesn't make sense. how can vt = -1/2gt^2. Since the ball always has an acceleration, there is no initial velocity which is what the v stands for. what i mean is that vt=0...... thats actually a really hard question. id be mad if they put that on the mcat. if you want to know how to do it, ask.

Also, initial v is not zero. the ball is thrown up!

I also want to point out that since this is a quadratic, you do get two answers: t = 0 and t = whatever time when ball hits the ground after being thrown. This makes physical sense because the ball is at its original height at both of these times, but we already know where the ball is when t = 0, so I ignored it.

 

[jerrywheat]

semantics....my point is that you only use vt when there is a constant velocity. With this problem there is no constant velocity, so vt=0(you remove it from the equation)

 

[FluffyRabbit]

The amount of time taken for the ball to reach the peak height is just v/g, and the amount of time taken for it to fall back down is exactly that as well. If g is 6 times smaller, it takes 6 times longer for it to reach the stationary point, and 6 times longer to fall. I never thought about the distance traveled at all.

 

[Isoprop]

semantics....my point is that you only use vt when there is a constant velocity. With this problem there is no constant velocity, so vt=0(you remove it from the equation)

I'm sorry, I made this unclear.

One of the equations for uniform acceleration (not constant velocity) is

y(t) = v(i)t +1/2at^2

v(i) is initial velocity. You're right that there is no one velocity during the trip. But there is an instantaneous velocity (which is always changing), and the initial velocity, v(i), is the velocity at the instant t=0. In this equation, it is a constant.

So when y(t) = 0

v(i)t = -1/2at^2

V(i)t does not equal 0.

 

[Isoprop]

The amount of time taken for the ball to reach the peak height is just v/g, and the amount of time taken for it to fall back down is exactly that as well. If g is 6 times smaller, it takes 6 times longer for it to reach the stationary point, and 6 times longer to fall. I never thought about the distance traveled at all.

Good way of thinking about it.

 

[jerrywheat]

sorry, just started physics...thought i knew what i was doing

 

[jerrywheat]

this is a completely different problem but how would you solve this: say a car is going at constant velocity for 8 seconds and then starts to accelerate at a constant acceleration for say 8 seconds?

 

[Isoprop]

For the first 8 seconds, you can model the problem with the equation

d = vt

Then, for the remaining 8 seconds, you can model the car using one of the uniform acceleration equations.

The initial velocity is the same velocity as the first equation. So if you wanted to find the total displacement of the car, and you knew the acceleration:

d1 = distance for the first 8 sec
d1 = vt
d1 = v * (8 sec)

d2 = distance for second 8 seconds
d2 = vt + 1/2at^2
d2 = v*(8 sec) + 1/2a * (8 sec)^2

Total displacement = d1 + d2

The v for both d1 and d2 is the same value. I just want to stress that the initial velocity of the car during acceleration is the same as the velocity of the car during the first 8 seconds.

BTW: I use d for displacement, which is equal to (x-xo) in the list of kinematics equations from the link.

 

[jerrywheat]

oh okay, thanks