# B Need help with gravitational acceleration SI units

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1. Jul 7, 2017

### icandothemath

Please ignore strikeout (not sure why post is doing that)
I just watched a video from Caltech (Video link at bottom of post)
It says:
Displacement = Distance Fallen in first second of time in [m] * Time2
so Time2 has no si unit as it is just a ratio right?

Instant or Average Velocity = 2 * Distance Fallen in first second of time[m] * Initial Time[sec] - Distance Fallen in first second of time[m] * Change in Time[sec]
velocity should have the si unit m/s but I am getting m*s (What am I doing wrong?)

Instant or Average Acceleration = 2 * Distance Fallen in first second of time[m] * (Initial Time[sec] + Change in Time[sec])
units should be m/s2 but I am getting m*s (What am I doing wrong?)

Gravitational Acceleration = 2 * Distance Fallen in first second of time[m]
units should be m/s[SUP]2[/SUP] but I am getting m (What am I doing wrong?)

Last edited: Jul 7, 2017
2. Jul 7, 2017

### pixel

Since this is a 28 minute video, can you tell us the time that this statement appears. The SI unit of time is seconds.

3. Jul 7, 2017

### Bandersnatch

When they talk about this (at around 12:30) mark, they specifically state that in the equation $s(t)=ct^2$, the constant c is numerically equal to the distance fallen in the first second. It does not mean that the constant is a measure of distance, or that its units are that of distance. It merely means that its magnitude is that of the distance fallen during the first second.
The dimensions of c can be deduced from the equation provided above:
if s(t) (distance) is measured in metres, and
t (time) is measured in seconds,
then the dimension of the constant must be [m/s^2], so that when you write the equation in terms of units, you get:
$m= m/s^2*s^2$
and the left side is equal to the right side.
That is, the constant c is acceleration.

Replace each instance of 'Distance Fallen in first second' with acceleration, and you'll have everything agree - everything apart from
Which should read $a(t) = 2c$, as they show in the video. (i.e., you made a mistake there)

4. Jul 7, 2017

### icandothemath

Thanks Bandersnatch,
Distance Fallen in first second = acceleration was the key to making everything work.

Displacement in meters = m/s2 * s2
Velocity in meters per second = 2 * m/s2 * s - m/s2 * s
Acceleration in meters per second squared = 2 * m/s2
Gravity in meters per second squared = 2 * m/s2

"Distance Fallen in first second" sounded like m/s (or velocity) at first for me. Like it falls this distance in a second.
How do you come up with acceleration?

5. Jul 7, 2017

### Bandersnatch

It does fall that much in the first second, but first second only. When you have $s(t)=ct^2$, and you set t=1s, then whatever is on the right hand side will tell you 's(1)', i.e. how much something falls in the first second. Since t=1, the only numerical value of this 'how much' must come from the constant c. But since setting t=1 did not make its units disappear (t^2 is 1 [s^2]), the dimensions of c can not be just metres. In fact, in order for the right hand side ([units of c] times [s^2]) to equal the left hand side ([m]), the dimensions of c must be m/s^2.

Imagine you have an object falling in gravitational field near the surface of Earth, with gravitational acceleration g=10m/s^2 (and c=g/2). The equation $s(t)=ct^2$ tells you that after one second, the object falling under this acceleration will have fallen 5 metres.

6. Jul 7, 2017

### icandothemath

The math you posted makes sense if we are really talking about "Velocity in the first second" m/s^2
I think the mess up in my mind is the description of "Distance Fallen in first second"

Should the speaker have said "Velocity in the first second"?

7. Jul 7, 2017

### Bandersnatch

No. Velocity after 1 second is V(1) = 2ct, with t=1s.

Analogous to the distance example, if you substitute t=1s, the V(1) will be numerically equal to 2c. In this case, the speaker would be entirely justified in saying that 2c has a magnitude of velocity after 1 second, even though c is still acceleration.
Dimensional analysis again proves that c is acceleration, since:
left hand side is [m/s]
right hand side is [units of c] times [s)
For both sides to be equal, [units of c] must be [m/s^2]

8. Jul 7, 2017

### icandothemath

Thanks for bearing with me, I don't know why this is so hard for me to get my head around. I think it is starting to sink in.
c = Acceleration = the numerical value of Distance Fallen in first second
So, we're really talking about acceleration in that first second.

At one second Distance, Velocity, and Acceleration are all the same numerical value but with different si units.
For example:
change in distance = d2 - d1 = 10m
change in velocity = change in distance / 1s = 10m/s (since initial velocity is 0)
Acceleration = change in velocity / 1s = 10/m/s^2 (since initial velocity is 0)

So, when he says Distance Fallen in the first second the numerical value works.
But we're really talking about Acceleration which makes the si units and numerical values work.