[Coda]

So the part that's confusing you -- and a small mistake Serra made -- is that the formula is specifically x = v_meant. It's not supposed to be the initial speed for the whole distance, it's the average speed.

For uniform acceleration, you should remember this from last semester:
v_mean = (v₀ + v_f)/2.

So Serra's math is off by a factor of 2, because the average speed during braking is not 88 ft/sec, but 44 ft/sec.

The rest of the process is correct:

x = v_meant
146 ft = (44 ft/sec) * t
t = (146 ft) / (44 ft/sec) ~= 3.32 sec

a = Δv/t
a = (-88 ft/sec) / (3.32 sec) -- this one is still 88 because it's the total change in velocity, not the average
a = -26.5 ft/sec²

EDIT: If you want to use v_f² = v_i² + 2a(x_f-x_i) you can, and it gives you the same result:

0 = (88 ft/sec)² + 2a(146 ft)
-7744 ft²/sec² = 2a(146 ft)
-53.0 ft/sec² = 2a
-26.5 ft/sec² = a

EDIT 2: Also your initial answer of 6.50x10⁴ mi/h² was in fact correct:

(65000 mi/h²) / (3600 sec/h) / (3600 sec/h) * (5280 ft/mi) = 26.48 ft/sec²

Dimensional analysis is a wonderful thing. :P

[Serra Britt]

Thanks for catching that Coda. I did miss it.

Also, I need to learn how to write sub and superscript like you XD

[Potironette]

Thank-you for explaining! I hadn't realized I was given all the stuff needed to find average velocity :o

Just [ s u b ][/ s u b] and [s u p][/ s u p] xD? (I forgot this existed, and I was lazy)

Also, one more question: for something that uses x_f = x_i + v_it + (1/2)at², if I just try to solve for t I get two answers. Usually one's negative and one's positive, but sometimes both are positive and in either case I don't know what they mean.

(More specifically, I have a problem that says: A boulder takes 1.30 s to fall the last 1/3 of the way down a cliff, and I have to (a)find the height of the cliff and (b)explain what the two solutions to a quadratic meant if I got two solutions and used a quadratic equation.

Along the way, I suppose I had to find the time too. (I didn't figure out how to do the problem myself so I basically looked at the solution and tried to understand it T_T)

Using x_f = x_i + v_it + (1/2)at², I get that
h = (1/2)gt²
(2/3)h = (1/2)g(t-1.30s)² --> h = (3/4)g(t-1.30s)²
(1/2)gt² = (3/4)g(t-1.30s)²

t² = (3/2)(t-1.30s)²
t = .716 s or t = 7.08 s

If I square rooted that above equation, the t = .716 s would disappear, but otherwise I would only know that t must be 7.08 s because .716 s is too short. But what does .716 s mean?)

[Coda]

You will always get two answers for a quadratic (that is, containing x²) equation unless the apex of the parabola is exactly on the x axis.

Do you already understand the physical significance if one value for t is negative? There's a meaningful answer but I don't want to do your homework for you.

If both values for t are in the future, there's a good reason you're confused: it PROBABLY means you've made a mistake somewhere. Either you're not considering the ground to be 0, or you've got the sign for one of the initial parameters wrong (the initial position, the initial velocity, or the acceleration), or you have an error in your computation somewhere, or you have one of the parameters wrong.

In this case, you've got the parameters wrong. Look at the formula more carefully:

x_f = x_i + v_it + (1/2)at²

I'll give you three hints:

1. If the height of the cliff is h, then what's the position of the boulder when you start looking at it?
2. What's the value of v_i?
3. You don't have enough information to solve this all with a single equation.

[Potironette]

I solved h to be 246 m based on the time being 7.08 s (since it couldn't have been .716 s because the total time was over 1.30 s).

But what I'm uncertain about is what the .716 s means. Although, I guess maybe .716 s might actually not in the future because when I solved for t, I got it from:
t² = 3/2(t-1.30 s)², and .716 s - 1.30 s < 0 ? Maybe since 7.08 s - 1.30 s is the time it took for the boulder to fall 2/3rds of the way, .716 s - 1.30 s is how long it would have taken the boulder to get to that same 2/3rds mark had the boulder been allowed to fall through the ground 2/3rds(I'm not sure exactly at what point) more then have time be rewound so the boulder would look like it was falling up from through the ground?? (All I remember from class about negative times was something about falling differently...granted this wasn't exactly a negative time, but even if I had considered the ground to be zero, I think the negative numbers would have canceled each other out?)

[Coda]

So yeah, you're still trying to solve it the same way. Try a different approach.

You know that it started at height 3h at time 0, and at time t₁ it's at height h, and at time (t₁ + 1.30). You're assuming that its velocity at time 0 is also 0, and the acceleration is a constant g, because they didn't say otherwise.

How can you plug these values into equations you know to get information you don't know?

Note that you shouldn't get two positive answers if you do this right because the top of the parabola is defined by these initial conditions to be at t=0.