Velocity of freefalling bodies after 1s, 2s, 3s, etc

[paulb203]

TL;DR Summary

How do v=at and v=Δx/t relate to each other when it comes to freefalling objects?

On many websites etc the velocity of a freefalling object in a vacuum is shown as follows.

After 1,2,3 and 4 seconds respectively;

9.8m/s; 19.6m/s; 29.4m/s; 39.2m/s

I worked out the distance travelled by a freefalling object in a vacuum using d = at^2/2, or, d = 0.5gt^2 and got, for after 1,2,3 and 4 seconds;

4.9m; 19.6m; 44.1m; 78.4m

But when I tried to double check the above using the formula for velocity, v= ΔX/t, I got, for after 1,2,3 and 4 seconds;

4.9m/s; 9.8m/s; 14.7m/s; 19.6m/s

Where have I gone wrong?

I’m always trying to bear in mind average velocity versus instantaneous velocity and I’m aware that there are several formulas for velocity but I thought those were mostly rearrangements of each other.

I see though that v=at works for the velocities at the top of this page.

How do v= Δx/t and v=at relate to each other?

 

[kuruman]

TL;DR Summary: How do v=at and v=Δx/t relate to each other when it comes to freefalling objects?

Where have I gone wrong?

v=Δx/t does not work in the case of accelerating objects because it implies that for constant v the object travels equal distances in equal times. When an object accelerates v is not constant.

 

[PeroK]

TL;DR Summary: How do v=at and v=Δx/t relate to each other when it comes to freefalling objects?

On many websites etc the velocity of a freefalling object in a vacuum is shown as follows.

After 1,2,3 and 4 seconds respectively;

9.8m/s; 19.6m/s; 29.4m/s; 39.2m/s

I worked out the distance travelled by a freefalling object in a vacuum using d = at^2/2, or, d = 0.5gt^2 and got, for after 1,2,3 and 4 seconds;

4.9m; 19.6m; 44.1m; 78.4m

But when I tried to double check the above using the formula for velocity, v= ΔX/t, I got, for after 1,2,3 and 4 seconds;

4.9m/s; 9.8m/s; 14.7m/s; 19.6m/s

Where have I gone wrong?

You've calculated the average velocity at each time - which, for constant acceleration, is half the velocity at that time.

In general :
$$v_{avg} = \frac{\Delta x}{t}$$

 

[Mister T]

TL;DR Summary: How do v=at and v=Δx/t relate to each other when it comes to freefalling objects?

In the first equation $v$ is the instantaneous velocity, but in the second equation $v$ is the average velocity. The two are not equal.

 

[jtbell]

But when I tried to double check the above using the formula for velocity, v= ΔX/t, I got, for after 1,2,3 and 4 seconds;

4.9m/s; 9.8m/s; 14.7m/s; 19.6m/s

These are the average velocities between the "time-points" 0 and 1s; 0 and 2s; 0 and 3s; 0 and 4s.

Notice that for the special case of constant acceleration, you can also calculate the average velocity as the average of the instantaneous velocities at the beginning and end of the time interval: $$\bar v = \frac {v_i + v_f} 2$$ For example, during the interval from 0s to 3s, $\bar v$ = (0m/s + 29.4m/s) / 2 = 14.7m/s, in agreement with your calculation.

 

[paulb203]

v=Δx/t does not work in the case of accelerating objects because it implies that for constant v the object travels equal distances in equal times. When an object accelerates v is not constant.

Thanks, kuruman.

So av.v=delta x/t works for, say, a car travelling 210 miles in 3 hours (the only information we are given)?

av.v=210mi/3hrs

av.v=70mi/hr

We know that accleration would have been involved but we’re only interested in the av.v in the above example?

 

[paulb203]

You've calculated the average velocity at each time - which, for constant acceleration, is half the velocity at that time.

In general :
$$v_{avg} = \frac{\Delta x}{t}$$

Thanks, peroK.

So the avg.v, when there’s constant acceleration, is half of the v at t=1s, t=2s, etc?

So after 1s the av.v is 4.9m/s, which is half of 9.8m/s...

After 2s the av.v is 9.8m/s, which is half of 19.8m/s...

After 3s the av.v is 14.7m/s, which is half of 29.4m/s...

And after 4s the av.v is 19.6m/s, which is half of 39.2m/s.

So, as Mister T said, with v=at we are talking about the INSTANTANEOUS velocity; but with v=delta x /t we are talking about the avg.v.

I’ve just split the first second (from 0s to 1s) into ten tenths and put g at 10m/s for simplicity. And got for v, after each tenth of a second, 1m/s, 2m/s, 3m/s, etc (using v = at, for the inst.v)

When I added those up and divided by 10 (for the average) I got 55/10=5.5. I’m guessing that if I did it more accurrately (putting g at 9.8m/s) I’d get 49/10=4.9. Do you think I would?

 

[paulb203]

These are the average velocities between the "time-points" 0 and 1s; 0 and 2s; 0 and 3s; 0 and 4s.

Notice that for the special case of constant acceleration, you can also calculate the average velocity as the average of the instantaneous velocities at the beginning and end of the time interval: $$\bar v = \frac {v_i + v_f} 2$$ For example, during the interval from 0s to 3s, $\bar v$ = (0m/s + 29.4m/s) / 2 = 14.7m/s, in agreement with your calculation.

Thanks, Mister T.

So at the end of the first second the INSTANTANEOUS velocity is 9.8m/s. At the end of the second second it’s 19.8m/s, etc, etc?

I remember hearing that INSTANTANEOUS velocity was actually either very difficult of even impossible to ascertain, given that when we do Speed = Distance/Time we can only get an average, even if we divide the distance into smaller and smaller units; but with a freefalling object, in a vacuum, can we safely say that after 1s we know it’s INSTANTANEOUS velocity to be 9.8m/s (or 9.81m/s; or whatever g is precisely)?

 

[paulb203]

These are the average velocities between the "time-points" 0 and 1s; 0 and 2s; 0 and 3s; 0 and 4s.

Notice that for the special case of constant acceleration, you can also calculate the average velocity as the average of the instantaneous velocities at the beginning and end of the time interval: $$\bar v = \frac {v_i + v_f} 2$$ For example, during the interval from 0s to 3s, $\bar v$ = (0m/s + 29.4m/s) / 2 = 14.7m/s, in agreement with your calculation.

Thanks, jtbell. It's beginning to make sense, I think, slowly though, as ever with me :)

 

[kuruman]

Thanks, kuruman.

So av.v=delta x/t works for, say, a car travelling 210 miles in 3 hours (the only information we are given)?

av.v=210mi/3hrs

av.v=70mi/hr

We know that accleration would have been involved but we’re only interested in the av.v in the above example?

It's not about what we are interested in. It's about what we can say. If all you know is that a car traveled 210 miles for 3 hours in a straight line, all you can say is that the car's average velocity is 70 mi/hr. That would be correct even if the driver stopped for a rest or moved in reverse for a while.

If in addition to that information if you are told that the acceleration is constant throughout the trip, then that information is sufficient to find the acceleration and from that where the car is at what time. Furthermore, when the acceleration is constant, the instantaneous velocity will be equal to the average velocity at the middle of the time interval. To be specific, in this case the car's speedometer will read 70 mi/hr when 1.5 hours have elapsed since it started moving from rest.

 

[jbriggs444]

I remember hearing that INSTANTANEOUS velocity was actually either very difficult of even impossible to ascertain, given that when we do Speed = Distance/Time we can only get an average, even if we divide the distance into smaller and smaller units

Yes. If you want to measure an instantaneous velocity, you have a difficult challenge.

You can try with a ruler and a stopwatch. But if you try to measure over increasingly smaller intervals, you will find that the markings on the ruler are too far apart and you will have problems stopping the stopwatch accurately.

So you switch to better instruments. You can now accurately assess a velocity over a much smaller time interval. Maybe you are now using a high speed video camera with megapixel resolution. But it is still not instantaneous.

So maybe you switch technology and use a laser interferometer (like the radar gun the police use, but better). But it is still not instantaneous.

Putting your engineering hat on, you realize that your instruments will always be imperfect. You can never measure an instantaneous velocity with perfect accuracy.

Putting your scientist hat on, you decide that there is nothing wrong with assuming that objects actually have instantaneous velocities and that these velocities change in a predictable way.

Putting your mathematician hat on, you come up with the notion of differential calculus. You understand that dividing zero incremental distance moved over zero incremental elapsed time is not a useful definition. So you decide to take the mathematical limit of the average velocity over an interval around the time of interest as the duration of that interval goes to zero.

Definition: (one of a variety of equivalent forms)$$v(t) = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t-\Delta t)}{2 \Delta t}$$

but with a freefalling object, in a vacuum, can we safely say that after 1s we know it’s INSTANTANEOUS velocity to be 9.8m/s (or 9.81m/s; or whatever g is precisely)?

Although we have not measured the instantaneous velocity and can never do so, every thing that we can measure is consistent with the Newtonian model. So we use it.

The Newtonian model assumes that things actually have instantaneous velocities that behave in the way that Calculus demands. The mathematicians assure us that this all makes sense.

[Most of] the things that we can measure accurately are consistent with the Newtonian model and tell us that the acceleration of gravity is approximately $9.8\ m/s^2$ near the surface of the earth.

 

[TonyStewart]

The exists a quadratic equation with the standard roots equation for solving time or distance given initial and final height, velocity for a constant acceleration.

 

[kuruman]

I remember hearing that INSTANTANEOUS velocity was actually either very difficult of even impossible to ascertain, $\dots$

@jbriggs444 makes the case that the instantaneous velocity is difficult to be measured and I agree. However, in your statement above you use "ascertain" which is different from "measure."

You put your scientist hat on and you measure the times at which the moving object is at fixed positions. You make a plot of position vs. time.
You put your mathematician hat on and you do a polynomial least-squares fit to the position vs. time $x(t)=a_0+a_1t+a_2t^2+\dots+a_nt^n~$ to get the coefficients to order $n$ beyond which there is no significant improvement to the fit.
You ascertain the instantaneous velocity to be $$v(t)=\sum_{k=1}^n ka_kt^{k-1}.$$The coefficients are used to interpolate the instantaneous velocity between the measured positions. One will need a model based on Newton's laws to attach physical meaning to them.

 

[paulb203]

It's not about what we are interested in. It's about what we can say. If all you know is that a car traveled 210 miles for 3 hours in a straight line, all you can say is that the car's average velocity is 70 mi/hr. That would be correct even if the driver stopped for a rest or moved in reverse for a while.

If in addition to that information if you are told that the acceleration is constant throughout the trip, then that information is sufficient to find the acceleration and from that where the car is at what time. Furthermore, when the acceleration is constant, the instantaneous velocity will be equal to the average velocity at the middle of the time interval. To be specific, in this case the car's speedometer will read 70 mi/hr when 1.5 hours have elapsed since it started moving from rest.

Thanks, kuruman.

So if we knew the acceleration was constant we could use the equation, v=at? By the way, should that equation have a letter, or letters, in subscript after the ‘a’ to indicate constant acceleration, as opposed to average acceleration, to avoid the confusion that I experienced? Something like ‘a’ followed by ‘c’ for constant, in subscript?

If we can use v=at we would get;

70mph = a x 3hr

a = 70mph / 3hr

a = 23 1/3 mph/h

Is that correct? Or does it have to be the standard units, m/s, s, m/s/s for these equations to work?

And, as in the initial freefall example, the v at 1 hr would be 23 1/3 mph; at 2 hr, 46 2/3 mph; and at 3 hr, 70mph?

Just like after 1,2 and 3 s in freefall we get 9.8m/s, 19.6m/s, 29.4m/s, yeah?

“... if acceleration is constant, the instantaneous velocity will be equal to the average velocity at the middle of the time interval. To be specific, in this case the car's speedometer will read 70 mi/hr when 1.5 hours have elapsed since it started moving from rest.”

Ah, this answers my above question. Where have I gone wrong?

 

[paulb203]

Yes. If you want to measure an instantaneous velocity, you have a difficult challenge.

You can try with a ruler and a stopwatch. But if you try to measure over increasingly smaller intervals, you will find that the markings on the ruler are too far apart and you will have problems stopping the stopwatch accurately.

So you switch to better instruments. You can now accurately assess a velocity over a much smaller time interval. Maybe you are now using a high speed video camera with megapixel resolution. But it is still not instantaneous.

So maybe you switch technology and use a laser interferometer (like the radar gun the police use, but better). But it is still not instantaneous.

Putting your engineering hat on, you realize that your instruments will always be imperfect. You can never measure an instantaneous velocity with perfect accuracy.

Putting your scientist hat on, you decide that there is nothing wrong with assuming that objects actually have instantaneous velocities and that these velocities change in a predictable way.

Putting your mathematician hat on, you come up with the notion of differential calculus. You understand that dividing zero incremental distance moved over zero incremental elapsed time is not a useful definition. So you decide to take the mathematical limit of the average velocity over an interval around the time of interest as the duration of that interval goes to zero.

Definition: (one of a variety of equivalent forms)$$v(t) = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t-\Delta t)}{2 \Delta t}$$

Although we have not measured the instantaneous velocity and can never do so, every thing that we can measure is consistent with the Newtonian model. So we use it.

The Newtonian model assumes that things actually have instantaneous velocities that behave in the way that Calculus demands. The mathematicians assure us that this all makes sense.

[Most of] the things that we can measure accurately are consistent with the Newtonian model and tell us that the acceleration of gravity is approximately $9.8\ m/s^2$ near the surface of the earth.

Thanks, jbriggs.

Your calculus equations looks fascinating, but beyond me at the moment; introductory maths and physics is a big enough challenge for me at the moment.

“...you decide that there is nothing wrong with assuming that objects actually have instantaneous velocities...”

Is there a hint of an implication there that assuming things have instantaneous velocities is controversial?

 

[PeroK]

Is there a hint of an implication there that assuming things have instantaneous velocities is controversial?

it's not controversial. The Newtonian equations of motion represents a mathematical model of particle motion. It's pointless, as far as physics is concerned, to debate whether they "really" do obey these laws. That the mathematical model produces accurate and useful predictions is enough.

Note that Newton's laws and classical trajectories are largely if not entirely abandoned in quantum mechanics.

 

[TonyStewart]

The quadratic equation for distance in the vertical axis with up=+ is ;
y = -0.5gt^2 + v₀t + h₀

Using the standard equation for computing roots of the quadratic equation of t, solve for t.
t = (-v₀ ± √(v₀² - 2gy₀)) / (-g)

• y is the height at time t
• g is the acceleration due to gravity ~ 9.8 m/s^2
• v₀ is the initial velocity
• h₀ is the initial height

 

[paulb203]

it's not controversial. The Newtonian equations of motion represents a mathematical model of particle motion. It's pointless, as far as physics is concerned, to debate whether they "really" do obey these laws. That the mathematical model produces accurate and useful predictions is enough.

Note that Newton's laws and classical trajectories are largely if not entirely abandoned in quantum mechanics.

Is that one for philosophers, or physicists who are also interested in philosophy, like Sean Carroll, for example, who I get the impression is quite into philosophy?

“Note that Newton's laws and classical trajectories are largely if not entirely abandoned in quantum mechanics.”

Is this a case of Newton not being wrong as such, but his theories etc only working up to a point, like when general relativity came along (I don’t pretend to understand much of that, but I have heard, roughly, that Einstein, building on the work of others, put forward the idea that gravity wasn’t a force that worked at a distance, rather space was curved, etc)?

 

[PeroK]

Is that one for philosophers, or physicists who are also interested in philosophy, like Sean Carroll, for example, who I get the impression is quite into philosophy?

“Note that Newton's laws and classical trajectories are largely if not entirely abandoned in quantum mechanics.”

Is this a case of Newton not being wrong as such, but his theories etc only working up to a point, like when general relativity came along (I don’t pretend to understand much of that, but I have heard, roughly, that Einstein, building on the work of others, put forward the idea that gravity wasn’t a force that worked at a distance, rather space was curved, etc)?

The important point is that your SUVAT formulas are a mathematical model with a certain range of applicability. They do not hold universally and shouldn't be seen as an objective reality in themselves.