I know that gravity is currently accelerating me at 32 feet/sec[sup]2[/sup], but let’s say I leap off my balcony on the 3rd floor. At the end of one second, I’m traveling at 32 feet per second. But I started at 0 feet per second.
Do I just average the two and come up with 16 feet in one second?
(Let’s disregard wind resistance, and I think better in feet/inches than cm/meters.)
Yes. In general distance travelled is 0.5 x a x t[sup]2[/sup] for constant acceleration (assuming a consistent set of units).
Yep, you got it. When dealing with constant acceleration from a stationary position, the average velocity is simply 1/2 the final velocity. Of course, the acceleration due to gravity is not constant, but diminishes the farther one gets from the center, but for everyday purposes, it is sufficient to assume that it is constant.
Thanks to both of you.
But you’d be falling more slowly at the halfpoint (time wise), so you’d have covered less than half the distance.
I agree. Its not a linear acceleration. I think we need a little calculus on this one.
In fact, it was for this very reason (to determine a falling object’s velocity at a specific point in the fall) that Sir Isaac Newton invented calculus. And then put it in a drawer and forgot about it for twenty years.
… Yeah, it is linear acceleration. You might need calculus to find the specific velocity at one point in time, but not the distance it travels in a given time. It’s a very simple kinematics problem.
x = x[sub]0[/sub] + v[sub]0[/sub]t + 1/2at[sup]2[/sup]
x = 0 + 0 + (1/2)(32)(1) = 16 ft
And I’m getting straight As in Physics. This is a very very simple Physics problem.
Q.E.D. and hibernicus are both correct (and Strinka too, on preview); to be honest, I’m not sure what you and the subsequent posters are referring to.
“…so you’d have covered less than half the distance.” Less than half the distance of what? The only question was how far you’d fall in one second; I’m not sure what it means to say “you’d have covered less than half the distance” here.
Maybe you’re thinking that you’d fall less in the first second than you will in the 2nd second. That’s certainly true: In one second, you’ll fall 16 feet, in a total of two seconds you’ll fall 64 feet. So yeah, in the first second you’ll fall less than half of the full 64 feet you will fall in the full two seconds, but that’s really not relevant to the question as it was asked, is it?
They may be referring to what I alluded to: that the Earth’s gravitational field strength drops off as the square of the distance from the center of mass. Therefore, the acceleration is slightly smaller at a given elevation than it is, say, 16 feet lower. However, given that the Earth has a radius of about 21,000,000 feet, the difference in acceleration over such a short fall is neglibible.
So let’s be clear. Yes, you will have fallen 16 feet — assuming away air resistance as you suggest, and assuming a constant gravitational field — but no, you don’t get that result by any averaging. It just “happens” to be the case that after 1 second, your speed is 32 feet/second, and your distance traveled is 16 feet, and that the number 16 is the average of 0 and 32. But that doesn’t hold in general.
Refering to the equation Strinka posted: after 2 seconds, your speed is 64 feet/sec, and your distance fallen is 64 feet (not 32). After 3 seconds, your speed isi 96 feet/sec, and your distance fallen is 144 feet (not 48). And so on. For this particular, simple problem, speed will be proportional to time, and distance will be proportional to the square of time.
“Half the distance you will have fallen after the full second is completed.”
Ahh, yes, If you really want to make it complicated… There are small variations from g due to wind resistance:
[a = g - b(v^2)/mass ] where b is a constant relating wind resistance to cross-sectional area.
(1) You can deal with it iteratively via a computer:
a = g - bvv/m
v = v + adt
x = x + vdt+(1/2)adt*dt
*Take dt as small as possible without waiting 10 years.
(2) Or you can roll out the Physical MacLaurin Series and
x = x_0 + v_0*t + (1/2)*a_0*t*t + (1/6)*(da_0/dt)*t*t*t + ...
and REALLY get down & dirty with the math, by hand.
My recommendation is just to blow off the 10^-4 order decimal points.
Actually, this problem is exactly soluble in one dimension, without any numerical evolution required:
v = sqrt(mg/b) tanh (sqrt(gb/m) (t - t[sub]0[/sub])),
and so
x - x[sub]0[/sub] = m/b ln(cosh(sqrt(m/gb)*(t - t[sub]0[/sub]))
where x[sub]0[/sub] and t[sub]0[/sub] are constants chosen so that the initial conditions are satisfied.
Oops, sorry, I just now realized I misread lissener’s post. I thought s/he was trying to find a flaw in the previous posts; now that I reread it, I realize s/he just meant the post in addition to what’d already been posted.
Yes, it does.
In uniformly-accelerated linear motion, you can calculate the distance travelled in a given time, t, by multiplying the average of the initial and final speeds by t.
Your calculations show this. The average of 64 (final velocity after 2 seconds) and 0 (initial velocity) is 32: 32 ft/s for 2 seconds gives you 64 feet. The average of 96 and 0 is 48: 48 ft/s for three seconds gives you 144 feet.
How far does the falling body move between the second and third seconds?
Initial velocity: 64 ft/s
Final velocity: 96 ft/s
Average velocity: (64 + 96) / 2 = 80 ft/s
Time: 1s
Distance travelled: 80 * 1 = 80 ft.
It’s constant acceleration (1 G) not linear acceleration (1G, 2G, 3G, …).
Right. The average velocity is vt/2 so the distance is vt/2 * t = vt[sup]2[/sup]/2 if you start with v[sub]0[/sub] = 0. I’m not sure it works for any other initial velocity.
lissener is a he.
It doesn’t work with other starting velocities. In that case the average velocity would be (v[sub]0[/sub] + at)/2 and the distance would be v[sub]0[/sub]t/2 + at[sup]2[/sup]/2. The right distance is v[sub]0[/sub]t + at[sup]2[/sup]/2.